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AN 



ELEMENTARY TREATISE 



CURVES, FUNCTIONS, AND FORCES. 



volume first; 



CONTAINING 



ANALYTIC GEOMETRY 
AND THE DIFFERENTIAL CALCULUS. 



By BENJAMIN PEIRCE, A. M., 

University Professor of Mathematics and Natural Philosophy in Harvard University. 



BOSTON: 
JAMES MUNROE AND COMPANY 

M DCCC XLI. 



■orb 



rii»ei«oi Y 



Entered according to act of Congress, in the year 1841, by James Munroe and Com- 
pany, in the Clerk's Office of the District Court of the District of Massachusetts. 



?> 



g-' ^ 






CONTENTS. 

BOOK I. 

APPLICATION OF ALGEBRA TO GEOMETRY, 
CHAPTER I. 

GEOMETRICAL CONSTRUCTION OF ALGEBRAICAL QUANTI- 
TIES 1 

CHAPTER II. 

ANALYSIS OF DETERMINATE PROBLEMS ... 18 

CHAPTER III. 

POSITION ......... 28 

CHAPTER IV. 

EQUATIONS OF LOCI ...... 52 

CHAPTER V. 

CLASSIFICATION AND CONSTRUCTION OF LOCI , . S7 

CHAPTER VI. 

EQUATION OF THE FIRST DEGREE .... 90 

CHAPTER VII. 

EQUATION OF THE SECOND DEGREE . . . 108 



p 



IV CONTENTS. 

I 

CHAPTER VIII. 

SIMILAR CURVES 140 

CHAPTER IX. 

PLANE SECTIONS OF SURFACES .... 143 



BOOK II. 

DIFFERENTIAL CALCULUS. 
CHAPTER I. 

FUNCTIONS . . 163 

CHAPTER II. 

INFINITESIMALS 172 

CHAPTER III. 

DIFFERENTIALS . . •'."'.'• • • • 1?9 

CHAPTER IV. 

COMPOUND AND ALGEBRAIC FUNCTIONS . . . 190 

CHAPTER V. 

LOGARITHMIC FUNCTIONS ] 96 

CHAPTER VI. 

CIRCULAR FUNCTIONS 200 

CHAPTER VII. 

INDETERMINATE FORMS 207 



CONTENTS. V 

CHAPTER VIII. 

MAXIMA AND MINIMA . . . . . . 213 

CHAPTER IX. 

CONTACT . . . . . . . . 220 

CHAPTER X. 

CURVATURE ........ 240 

CHAPTER XL 

SINGULAR POINTS 248 

CHAPTER XII. 

APPROXIMATION ....... 295 



BOOK I. 



APPLICATIOiN OF ALGEBRA TO GEOMETRY. 



CURVES AND FUNCTIONS. 



BOOK I. 

APPLICATION OF ALGEBRA TO GEOMETRY. 

CHAPTER I. 

GEOMETRICAL CONSTRUCTION OF ALGEBRAICAL QUANTI- 
TIES. 

1. In the application of Algebra to Geometry, usually 
called Analytic Geometry, the magnitudes of lines, 
angles, surfaces, and solids are expressed by means of 
letters of the alphabet: and each problem, being put 
into equations by the exercise of ingenuity, is solved 
by the ordinary processes of Algebra. The algebraical 
result is finally to be interpreted geometrically : and 
this geometrical interpretation of an algebraical expres- 
sion is called the geometrical construction of that ex- 
pression. The geometrical construction of the results 
is, then, the last operation in the solution of problems j 
but it is convenient, on account of its simplicity, to 
begin with the consideration of it. We begin with 
the easiest cases and proceed to the more difficult ones, 
and we regard each letter as representing a line, so that 
1 



2 ANALYTICAL GEOMETRY. [fi. I. CH. I. 

Sum and difference. Negative sign. 

every algebraical expression of the first degree will de- 
note a line ; whence it is called linear. 

2. Problem. To construct a -\- b. 
Solution. Take (fig. 1.) 

AB = a, 

BC=b; 
and we have 

AC=AB + BC=a + b; 
so that AC is the required value of a -|- b. 

3. Problem. To construct a — b. 

Solution. Take (fig. 2.) 

AB = a, 
and from JB, in the opposite direction, 

BC=b; 
we have then 

AC = AB — BC = a — b ; 
so that AC is the required result. 

4. Corollary. If a were zero, the preceding solution would 
become the same as to take from A (fig. 3.) in the direction 
AC, opposite to AB, 

AC=b; 
so that the negative sign would only be indicated by the 
direction of AC. In order to generalize the preceding con- 
struction we must, then, adopt the rule that 

The geometrical interpretation of the negative sign 
is opposite direction. 

5. Problem. To construct an algebraic expression con- 
sisting of a series of letters connected together by the 
signs -{- and — . 



<§> 9.] GEOMETRICAL CONSTRUCTION. 3 

Product and Quotient. Surface and Solid. 

Solution. Collect into one sum, by art. 2, all the letters 
preceded by +, which sum may be denoted by a ; and col- 
lect into another sum all the letters preceded by — , which 
sum may be denoted by b ; and the value of a — b may then 
be constructed by art. 3. 

6. Corollary, If a letter is preceded by an integral 
numerical coefficient, it may be regarded as a letter re- 
peated a number of times equal to this integer. 

7. Problem. To construct a b. 

Solution. The parallelogram of which the base is a, and 
the altitude is b, is equal to the product a b, which accordingly 
represents a surface ; and this conclusion is a general one, 
that is, 

A homogeneous algebraical expression of the second 
degree represents a surface. 

8. Problem. To construct a b c. 

Solution. The parallelopiped of which the base is the par- 
allelogram a b, and the altitude is c, is equal to the product 
a b c, which accordingly represents a solid ; and, in general, 

A homogeneous algebraical expression of the third 
degree represents a solid. 

9. Problem. To construct t- 

b 

Solution. Make (fig. 4.) the right angle ABC, 
take AB = a 

BC = b, 
and join AC. The angle ACB is, by trigonometry, that 



angle whose tangent is - 



4 ANALYTICAL GEOMETRY. [b. I. CH. I. 

Angle. To render homogeneous. 

10. Corollary. If we had taken 

AC= b, 

the angle ACB would have been the angle, whose sine is -, 

and, in general, 

A homogeneous algebraical expression, whose degree 
is zero, represents the sine, tangent, fyc. of an angle. 

11. Scholium. Since no other magnitudes occur in 
Geometry but angles, lines, surfaces, and solids, all 
algebraical quantities which represent geometrical mag- 
nitudes must be either of the 1st, the 2d, the 3d, or the 
zero degree ; and since dissimilar geometrical magnitudes 
can neither be added together, nor subtracted from each 
other, these algebraical expressions must also be homo- 
geneous. 

If, therefore, an algebraical result is obtained, which 
is not homogeneous, or is of a different degree from 
those just enumerated ; it can only arise from the cir- 
cumstance, that the geometrical unit of length, being 
represented algebraically by 1, disappears from alf alge- 
braical expressions in which it is either a factor or a 
divisor. To render these results homogeneous, then, 
and of any required degree, it is only necessary to re- 
store this divisor or factor which represents unity. 

12. Problem. To render a given algebraical expres- 
sion homogeneous and of any required degree. 

Solution. Introduce 1, as a factor or divisor, repeated 
as many times as may be necessary, into every term 
where it is required. 



§ 14.] GEOMETRICAL CONSTRUCTION. 

To render homogeneous of any degree. 



13. Examples. 

1. Render *T "J* — homogeneous of the 1st degree. 

Ans . »»* + (!)' « + !•<*», 
l.e -f- h 2 

a -4- d 3 -f- h e 

2. Render — ~ — — — homogeneous of the 2d degree. 

I 2 -\- m 3 b 5 

(1)* a+(l) 2 rf3 + (l) 3 /ig 



^4ws. 



_}_ 6 c -J- d b h 2 
3. Render — — homogeneous of the 3d degree. 



Ans. 



1 . I 2 + m 3 
)us of the 3d 
(1)6 a + (l)sbc + d* h 2 



(i)2de — (l) 3 a 
4. Render — homogeneous of the zero degree. 



5. Render a b homogeneous of the 1st degree. 



Ans. 



1. c — d 2 ' 
th( 
ab 



6. Render ab c -\- d — e 2 homogeneous of the 1st degree. 

ab c 



-'"Hi + d ~ ' T 



14. Scholium. By the preceding process, every frac- 
tion, which does not involve radicals, may be reduced 
to a homogeneous form, in which each term is of the 
first degree ; and, although this form is not always that 
which leads to the most simple form of construction, 
its generality gives it a peculiar fitness for the general 
1* 



6 ANALYTICAL GEOMETRY. [b. I. CH. I. 

To render homogeneous. Fraction. 

purposes of instruction, where the artifices of ingenuity- 
are rather to be avoided than displayed. 



15. Examples. 

1. Reduce the fraction of example 1, art. 13, to a homo- 
geneous form, in which each term is of the first degree. 



-•(S'+'+f ■)+(*¥)■ 



2. Reduce the fraction of example 2, art. 13, to a homo- 
geneous form, in which each term is of the first degree. 

/ . dz , he \ i 12 . m * \ 
Ans . {a+-- 2 +T )-^ T +_). 

3. Reduce the fraction of example 3, art. 13, to a homo- 
geneous form, in which each term is of the first degree. 






4. Reduce the fraction of example 4, art. 13, to a homo- 
geneous form, in which each term is of the first degree. 



16. Problem. To construct — . 

c 



Solution. We have 



_ ab 
c : a = b : — , 
c 



that is, the given fraction is a fourth proportional to the three 
lines, c, a, and b. 

Find, then, by geometry, a fourth proportional to the 



<§> 19.] GEOMETRICAL CONSTRUCTION. 7 

Monomial. 

lines c, a, and b ; and this fourth proportional is the 
required result. 

a 2 
17. Corollary. The value of — is a third proportional to 

c 

c and a. 

18. Problem. To construct any monomial which de- 
notes a line. 

Solution. If the monomial is not of the first degree, 
reduce it to the first degree by art. 12. It is then of 
the form 

abed... ab c d 

T~Ti '/ == 7 X ~~T, X r • • • • 

a' b' d . . . a 1 b c 

Construct first — 3 and let m be the line which it re- 

,—. . . \ m c 

presents. 1 he given quantity becomes — - X 



Let, again, m! = 



b> N c' 
m c 



m' d 
also m" = —-, &c. 

c 

and the last line thus obtained is plainly the required 
result. 



19. Examples. 

1. Construct the line ab. Ans. m = — = ab. 

a b mc 

2. Construct the line a be. Ans. m = — , m' — — = a b c. 



ANALYTICAL GEOMETRY. [b. I. CH. I. 



Any expression not involving radicals. 



_ _ , ,. .. a* , m a 

3. Construct the line a 5 . Ans. m = — , ml = —-, 

m'a m" a 

m" = — - — , m" =. — — = a 5 . 

4t. Construct the line 7o . Ans. m = — r , m' = — -- 
d 3 d' d 

m' . 1 . a 2 b 



d ~ d* 

2 ab 2a. b , m 
5. Construct the line — ■=—.. Ans. m = , m 



efg e ' '■ f 

1 2ab 



m T— • 

S e fg 

1 (I) 2 1 

6. Construct the line -. Ans. m = — — = -. 

a a a 

A 1 A 4. 

7. Construct the line — . Ans. m = * == ^—. 

JtS Jd H 

20. Corollary. By this process each term of an al- 
gebraic expression, which does not involve radicals, is 
reduced to a line ; and if the expression does not involve 
fractions, it may then be reduced to a single line by 
art. 5 ; if it does involve fractions, the numerator and 
denominator of each fraction is, by art. 5, reduced to a 

single line, and each fraction, being then of the for in -=* 

is constructed like example 7 of the preceding article, 
and the aggregate of the fractions is then reduced to 
a single line, by art. S. 

Any algebraic expression, which represents a line, 
and does not involve radicals, may therefore be con- 
structed by this process. 



<§> 21.] GEOMETRICAL CONSTRUCTION. 



Any expression free from radical: 



21. Examples. 



, _, , ,. a 2 b + c -f d 2 

1. Construct the line 



e -f h 2 
Solution. Let m = = a 2 b 

= t = d* 







1 




m" 


r= 


7*2 


-h* 


and the fraction becomes 








7?Z 


+ 


c + 


m! 



e -f- m" ' 
let now A — m -f- c -\- m' 

B — e + m", 

1 ..4 

and the line represented by — — - is the required line. 

2. Construct the line represented by the fraction of example 
2, art. 13. 

Ans. m — d 3 , m" — he, in'" = I 2 , ?n ir = m 3 , 
A = « + »»' -J- in", i? — m 1 " -\- m IV , 

and the required line is the fourth proportional to B, 1, 
and A. 

3. Construct the line represented by the fraction of exam- 
ple 3, art. 13. 

Ans. Let m = b c, m' = d 5 h 2 , m" -de, 
A = a -\- m -f- m', B = m" — a, 

and the required line is the fourth proportional to B, 1, 
and A. 



10 ANALYTICAL GEOMETRY. [b. I. CH. I. 

Radicals of the second degree. 

4. Construct the line represented by the fraction of exam- 
ple 4, art. 13. 

Ans. m == a 2 , m' = d 2 , 

A — m -|- 6, B =: c — m', 
and the required line is the fourth proportional to B, 1, 
and A. 

5. Construct the line represented by the polymonial of ex- 
ample 6, art. 13. 

Ans. ?n — abc, m' = e 2 , 

and ^L — m -f- b — m is the required line. 

22. Problem. To construct the line \/(a b). 

Solution. Since */{a b) is a mean proportional be- 
tween a and b, the required result is obtained by con- 
structing, geometrically, this mean proportional between 
a and b. 

23. Corollary. The expression 

s/A = */(!. A) 
may be constructed by rinding a mean proportional be- 
tween 1 and A. 

24. Corollary. The square root of any algebraical 
expression, which does not involve radicals, may be 
constructed by finding, as in art. Ji9, the line A, which 
this algebraic expression represents, and then construct- 
ing */ A as in the preceding article. 

By the repeated application of this process, any al- 
gebraic expression may be constructed which represents 
a line, and which does not involve any other radicals 
than those of the second degree. 



yc r 



§ 27.] GEOMETRICAL CONSTRUCTION. 11 

Radicals of the second degree. 



25. Examples. 

1. Construct the line \/(a -\- b — c — e.) 

Ans. A — a -\-b — c — e, 
and \/A is the required line. 

2. Construct the line \/(a 2 -f- am). 

Ans. A — a 2 -\- am, 
and a>/A is the required line. 

n „ ii- \/ab 4- \/(e f — h) 

3 Construct the line — ' .; v J r — '-. 

e 2 — v (c 2 — j n) 

Ans. m=z\Zab } m = \/(ef — h), ro" = e 2 , m'"—*/(c 2 — fn), 
A = m + m', B = m" — m' 1 ', 

and the line — is the required line. 

4 

4. Construct the line \/a. Ans. m = \Za, 

and \/m is the required line. 

26. Scholium. When the expression whose square 
root is required is easily decomposed into two factors, 
it is immediately reduced to the form */(ab) and con- 
structed, as in art. 22. 



27. Examples. 

1. Construct example 2, art. 25, by decomposing the quan- 
tity under the radical sign into two factors. 

Solution. a 2 -f- a m = a (a -f- m). 

Let b z=z a -f- m, 

and the line \/(a b) is the required line. 



12 ANALYTICAL GEOMETRY. [b. I. CH. I. 

Square root of sum and difference of squares. 

2. Construct \/(a 2 -\- ae -\- am — an) by decomposing 
the quantity under the radical sign into two factors. 

Arts, b = a -\- e -\- m — n, 
and \/(ab) is the required line. 

3. Construct \/(a -\- a 2 — a 3 ) by decomposing the quan- 
tity under the radical sign into two factors. 

Ans. b = 1 -\- a — a 2 , 
and \/(a b) is the required line. 

4. Construct s/ia 2 — b 2 ) by decomposing the quantity 
under the radical sign into factors. 

Ans. c =± a -f- b, e = a — b, 
and \/(c e) is the required line. 

28. Scholium. Example 4 of the preceding article 
may also be solved by constructing a right triangle, of 
which a is the hypotlienuse and b a leg, and*/(a 2 — b 2 ) 
will be the other leg. 

29. Corollary. In the same way \/(a 2 + ° 2 ) *? the 
hypotlienuse of a right triangle, of which a and b are 
the legs. 

30. Corollary. By combining the processes of the 
two preceding articles, any such expression as 

^/(«2 _|_ b2 _ C 2 _ e 2 _|_ ] L 2 _|_) ; & c# 

may be constructed. For if we take 

m — */(« 2 + 1> 2 ), m ~ /^{m 2 — c 2 ), 

m" = \Z(m' 2 — e 2 ), m"> — ^/(m" 2 + h 2 ), &c. 

we have m 2 = a 2 + b 2 , 

rri - \/(m 2 — c 2 ) = */{a 2 + b 2 — c 2 ), 

or m 12 — a 2 + b 2 — c 2 ; 



§ 33.] GEOMETRICAL CONSTRUCTION. 13 

Construction of radicals. 

m" = \/(m' 2 — e 2 ) =3 \/(a 2 + b 2 — c 2 — e 2 ), 
or m" 2 = a 2 + b 2 — c 2 — e 2 ; 

m'"=z^{rn" 2 + h 2 ) — ^/(a 2 + 6 2 — c 2 — e 2 + A 2 ), &c. 

31. Corollary. The square root of the sum or dif- 
ference of any expressions, which involve no other 
radicals than those of the secojid degree, may also be 
constructed by the preceding process. For if either of 
these expressions is constructed by the processes before 
given, it may be represented by A ; and, if we denote 
as/ A by m, we have 

ra 2 z± A, 

so that each expression is reduced to the form of a 
square, and the whole radical is reduced to the form of 
the preceding article. 

32. Examples. 

1. Construct example 1, art. 24, by the process of art. 30. 2-j 

Ans. in = \/a, w! =z s/b, m" = \/c, ml" c= \/e f 
and the line \/(m 2 -f- W 2 — m" 2 — m"' 2 ) is the required line. 

2. Construct example 2, art. ^24, by the process of art. 30. 

Ans. m! = \/(a in), and \/( a2 + m ' 2 ) is tne required 
line. 

3. Construct the line \Z(a 2 -\-bc — e 2 -\-h) by process of 
art. 30. 

Ans. m = s/(b c), m' — s/e 3 , m" = s/h, 
and the line \/(a 2 + w 2 — m' 2 -\-m" 2 ) is the required line. 

33. Problem. To construct an algebraical expression 
which represents a surface. 
2 



14 ANALYTICAL GEOMETRY. [b. I. CH. I. 

Surface. Solid. Angle. 

Solution, Let A be the line which is represented by 
this algebraical expression, and since we have 

A = 1.A, 

the required surface is represented in magnitude by the 
parallelogram, whose base is 1, and altitude A, or by 
the equivalent square, triangle, &c. 

34. Problem. To construct an algebraical expression 
which represents a solid. 

Solution. Let A be the line which is represented by 
this expression, and since we have 

the required solid is represented in magnitude by the 
parallelopiped, whose base is the square (I)' 2 , and whose 
altitude is A. 

35. Problem. To construct an algebraical expres- 
sion, which represents the si?ie : tangent, 6fc. of an 
angle. 

Solution. Let A be the line which is represented by 
this expression, and since we have 

A- A 

the required angle is found by art. 9 or 10. 

36. Scholium. The construction of all geometrical 
magnitudes being, by the three preceding articles, re- 
duced to that of the line ; we shall limit our construc- 
tions hereafter to that of the line. 



§ 38.] GEOMETRICAL CONSTRUCTION. 15 



Equations of first and second degree. 



37. Problem. To construct the root of an equation 
of the first degree with one unknown quantity. 

Solution. Every equation of the first degree may, 
as is proved in Algebra, be reduced to the form 

A x + M — j 
whence 

X ~ A' 

and this value of x may be constructed by art. 18. 

38. Problem. To construct the roots of an equation 
of the second degree with one unknown quantity. 

Solution. The equation of the second degree may, as 
is shown in Algebra, be reduced to the form 

A x 2 + B x + M ^= 0. 

If we divide this equation by A, and put 





B M 
a =2A> m =A> 




it becomes 








x 2 -{-2ax-{-m=. 


0. 


The roots of this last equation are 






x =. — a ^z \/(a 2 — 


m) 



Case 1. When m is positive and greater than a 2 , 
the roots are both imaginary, and cannot be con- 
structed. 

Case 2. When m is positive and equal to a 2 , each 
root is equal to — a, which needs no farther construc- 
tion. 



16 ANALYTICAL GEOMETRY. [b. I. CH. I. 

Quadratic equation. 

Case 3. When m is 'positive and less than a 2 . Let, 
in this case, b =s \^m, or b 2 =. m. 

The roots become 

x = — a ± \/(« 2 — b 2 ), 
which are thus constructed. 

Draw (fig. 5.) tfAe fa#o indefinite lines DAD' and 
AB perpendicular to each other. Take 

AB = b; 
from B as a centre, with a radius 

BC = a, 
describe an arc cutting DAD' in C. Take 

CD = CD' = BC = a, 
and the required roots, independently of their signs, are 
AD and AD'. 
Demonstration. For 

AC = a/{BC 2 — AB 2 ) — \/(a 2 — b 2 ) 
and — AD = — CD + A C — — a + */{a 2 — b*) 
— AD' — — CD' — AC——a — s/{a 2 — b 2 ). 

Case 4. When rn is zero, the roots are 
x = and x = —%a, 
which require no further construction. 

Case 5. When m is negative, so that — m is positive. 
Let b = \/ — m, or Z> 2 = — m. 

The roots become 

x = — a ± s/(a 2 + b 2 ), 
which are thus constructed. 



$ 39.] GEOMETRICAL CONSTRUCTION. 17 

Quadratic equation. 

Draw (fig. 6.) the two lines AB and AC peiyen- 
dicular to each other. Take 

AB = a, and AC = b ; 
through BC draw the indefinite line BCD'. Take 
BD = BD' = AB=a, 

and the required roots, independently of their signs, are 
CD and CD'. 

Demonstration. For 

BC— */(ABz + AC*) = V(a 2 + & 2 ) 
and 

CD — —BD + BC= — a + s/{a* + b*) 
— CD — — BD' — BC — — a — s/(a* + 6 2 ). 

39. Scholium. Radicals of a higher than the second 
degree, and roots of equations of a higher than the 
second degree, do not usually admit of geometrical 
construction. 



2* 



18 ANALYTICAL GEOMETRY. [b. I. CH. II. 

Solution of determinate problems. 



CHAPTER II. 

ANALYSIS OF DETERMINATE PROBLEMS. 

40. Geometrical problems are of two classes, deter- 
minate and indeterminate. 

Determinate problems are those, which lead to as 
many algebraical equations as unknown quantities ; and 
indeterminate problems are those, in which the number 
of equations is less than that of the unknown quanti- 
ties. 

41. The solution of a geometrical problem consists 
of these three parts ; 

First, the putting of the question into equations ; 
Secondly, the solution of these equations ; 
Thirdly, the geometrical construction of the algebra- 
ical results. 

The last of these processes has been treated of in the pre- 
ceding chapter, but it must be observed that much skill is 
often shown in arranging the construction in such a form, that 
it may be readily drawn and be neat in its appearance. 

The second process is exclusively algebraical, and the first 
process, the putting into equations, is a task which, necessarily, 
requires ingenuity, and can only be taught by examples. One 
great object is to obtain the simplest possible equations, and 
such as do not surpass the second degree. It is not unfre- 
quently the case, that, when a question admits of several solu- 
tions, two or more of these solutions are connected together in 






§ 41.] DETERMINATE PROBLEMS. 19 

Division of line. 

such a way, that the same quantity, being obviously common 
to them, should, on this account, be selected as the unknown 
quantity. 



42. Examples. 

1. To divide a line AB (fig. 7.) into two such parts, that 
the difference of the squares described upon the two parts may 
be equal to a given surface. 

Solution. Let the magnitude of the given surface be equal 
to that of the square whose side is AC, and let D be the 
point of division, AD being the greater part. Let 

a — AB, b =z AC, and x — AD, 
we have then 

BD = a — x\ 

and the equation for solution is 

X 2 __ ( a __a;)2 _ 12 . 

or 2 ax — a 2 = b 2 . 

b 2 + a 2 62 
Hence x =s — ^ z= f- A a. 

Construction. Let E be the middle of AE. Draw the in- 
definite line EB'. Take 

EB 1 — EB — \a, 

EC — EC' = iAC=ib. 

Join B 1 C, and through C" draw CD parallel to B C, D is 
the point of division required. 

Demonstration. We have 

EB 1 : EC = EC ; ED, 
or %a :lrb =%b : ED ; 



20 ANALYTICAL GEOMETRY. [b. I. CH. II. 



Rectangle inscribed in triangle. 



whence ED = \b 2 -±- J a = -—, 



&2 

2? 



and AD — ED + AE — ~- + J a. 

' 2<2 

2. To inscribe in a triangle ABC (fig. 8.), a rectangle 
DEFG whose base and altitude are in the given ratio m : n. 

Solution. Let fall the perpendicular AIH. Let 

BC — b, AH= h y 

DE = HI=z, AI= AH—HI—h — x; 

and, since n : m = Z>£ : JE^F, 

_ _, W2 X 

we have -Ei 1 =: — . 

n 

But the triangles AEF and ABC are similar, and their 

bases are, therefore, proportional to their altitudes, that is, 

BC: EF~ AH: AI, 

mx 

or o : — It : li — z. 

n 

Hence we find, by algebraical solution, 

n b h 



= n+m+i)i. 



m h-\-nb 

Construction. Find a fourth proportional to n, m, and h t 
and denote it by h', and then x is obviously a fourth propor- 
tional to h' -\-b, b and h. 

The following simple form has been obtained by geometers. 
Draw AK parallel to BC, and take 

AK — h'. 

Join KC } and ED is the required altitude. 



<§> 41.] DETERMINATE PROBLEMS. 21 

Line of given length intercepted between parallels. 



Demonstration. For since h', or its equal AK, is a fourth 
proportional to n, m, and h, we have 

n : m = h : AK, 
or AK : AH =z m : n. 

If we let fall the perpendicular KL upon BC, we have the 
quadrilateral CFED, CAKL, which are formed of similar 
triangles ; they are therefore similar, and their homologous 
sides give the proportion 

FE DE=z AK: KL (or AH) = m : n ; 
so that FE and DE are in the required ratio. 

Corollary. If the ratio m : n were that of equality, the 
rectangle would be a square. 

3. To draw through a given point A (fig. 9.) situated be- 
tween two given parallels BC and DE a line HI, which may 
be of a given length a. 

Solution. Since the point is given, its distances from the 
parallels must be given, which are 

AF=zb, AG~c% 
let AH—z) 

we shall leave it as an exercise for the learner to find the value 
of x, which is 

a b 

Construction. The value of 2; is a fourth proportional to 
b -\- c, a, and b, and may be easily constructed. 

The following form is quite simple. From G as a centre, 
with a radius equal to a, describe an arc cutting BC in K. 
Join GK, and the line drawn through A parallel to GK is 
obviously of the same length with GK, and it is, therefore, 
the required line. 



22 ANALYTICAL GEOMETRY. [b. I. CH. II. 

Circle tangent to given line. 



Corollary. The problem is impossible when the length a is 
less than GF or its equal b ~\- c. 

4. To draw a circle through two given points A and B 
(fig. 10.), and tangent to a given line DC. 

Solution. Join AB, and produce AB to meet DC at D. 
Let C be the point of contact, and let 

DA = a, DB =.b, DC— x> ; 

we have, by geometry, 

DA :DC=DC:DB, 

or a : x r= x : b ; 

whence x =± ± \/(ab). 

Construction. Find a mean proportional between a and b, 
and take DC or DC equal to it, and Cor C is the point of 
contact, these two values corresponding to the two different 
circles BCA and BCA. 

Instead of finding the mean proportional by the ordinary 
process, we may find it, by drawing any arc AEB through 
A and B, and' the tangent DE to this arc is, by geometry, 
the mean proportional between DA and DB. 

Corollary. The problem is impossible if a and b are of 
opposite signs, that is, if A and B are in opposite directions 
from D, one being above the line and the other below it. 

Corollary. If either a or b is zero, as in fig. 11, where 

DA = a rrr 0, 

the problem is reduced to that of finding a circle which passes 
through the given point B, and is tangent to a given line CA 
at a given point A. 

Construction of this case. Erect OA perpendicular to AC. 
Join AB, and at the middle E of AB erect the perpendicular 



$ 41.] DETERMINATE PROBLEMS. 23 

Division of a line. 



EO ; O is the centre. The demonstration of this construction 
is left as an exercise for the learner. 

Corollary. If a and b are equal, as in fig. 12, the problem 
becomes ; to find a circle which touches a given line DA at 
a point A, and also touches another given line DC. 

Construction. Take 

DC— DC — a, 
and the point O or O', the intersection of the perpendicular 
OAO' } with the perpendicular CO or CO', is the centre of 
the required circle. 

5. To divide a given line AB (fig. 13.) into two such parts, 
that the sum of the squares described upon the two parts may 
be equal to a given surface. 

Solution. Let the given surface be twice the square whose 
side is AC, and let D be the point of division. Let also E 
be the middle of the line, and let 

BE — AE — a, AC — b, DE 'i= x. 

The value of a; will be found to be 

x fc ± V(6 2 — a 2 ), 

so that z is a leg of a right triangle whose hypothenuse is b 
and other leg a. 

Corollary. The problem is impossible when b is less than a 
and also when 

%>a, 
or b 2 — a 2 > a 2 , 

or 6 2 >2a 2 , 

or 2b 2 >ia 2 , 

2 b 2 > (2 a) 2 ; 



24 



ANALYTICAL GEOMETRY. [b. I. CH. II. 



Line divided in extreme and mean ratio. 



that is, when the given surface is greater than the square of 
the given line. 

6. To divide a given line AB (fig. 14.) at the point C in 
extreme and mean ratio. 

Solution. Let AC be the greater part, and let 

AB == a, AC — x, CB — a — z, 

we are to have 

t a : x =z x : a — x, 

whence we find 

x 2 -\- ax — a 2 — 0, and x = £ a ( — 1 i \/5). 

Construction. The roots of the equation 



-j- a x 



= 0, 



being constructed by case 5, art. 38, give the usual construc- 
tion of this problem. 

7. Through a given point C (fig. 15.) to draw a line BCD, 
so that the surface of the triangle ABD intercepted between 
the lines AB and AD may be of a given magnitude. 

Solution. Let the given surface be double that of the given 
rhombus AEFG. Draw CH parallel to AD, and CI parallel 
to AB. Let 

AI— CH— a, AH— CI— b, 
AE = c, AD — x, AB — y. 
"We have 

surface of triangle = \xy sin. A = %c 2 sin. A, 



whence 



xy 



The similar triangles BHC, BAD, give 
BH.HC— BA.AD, 

or y — b:a = y:x; 

whence xy — ay + bx. 



§ 42.] DETERMINATE PROBLEMS. 25 

Given length intercepted. 
The solution of these equations gives 

x=:^ ( C ±\/(c 2 — ab)\ 

which are easily constructed. 

Corollary. The problem is impossible when c 2 is less than 
a b ; that is, when 

c 2 sin. A < a b sin. A, 

or when the rhombus AEFG is less than the parallelogram 
AHCL 

8. "Through a given point C (fig. 16.) to draw a line BCD, 
so that the part BCD intercepted between two given lines 
AB and AD may be of a given length, the point C being at 
equal distances from the two given lines. 

Solution. Draw CH and CI parallel respectively to AB 
and AD, and they are obviously equal to each other. Let 

then AH— AI — CH = CI — a, BD = b t 

AD — x, AB — y. 

From triangle ABD, we have 

x 2 -\- y 2 — 2 x y cos. A — b 2 ; 

and, from similar triangles BIC and BAD, 

x y = a (x + y). 

As these equations are symmetrical with regard to x and y, 
they are simplified by putting 

x + y — 5, .xy — t; 

3 



26 ANALYTICAL GEOMETRY. [b. I. CH. IT. 

Given length intercepted. 

and become t =. as 

s 2 — 2 a (1 + cos. A) s = b 2 ; 
whence s and t f x and y are found. 

The following solution is, however, much neater. Join AC, 
and let the angle ACD be the unknown quantity, and put 

c±zAC, CAD — ^A = A', ACD = <p, 

ADC = 180° — (c P + A') 9 ABC— <p — A', 

sin. ADC = sin. (<r + A') ; 

and, by trigonometry, 

sin. ADC : sin. DAC — AC : DC, 

sin. (<p + 4') : sin. A' = c : DC; 

c sin. A' 
Hence DC == - — - — ■ — — . 

sin.((/) -j-^L ) 

Also sin. ABC : sin. BAC — AC : BC, 

•sin. (<p — 4') : sin. A' = c : £C; 

c sin. ^4' 



£C = 



sin.(<p — A')' 



am I ir\ A I \ > 



sin. (9 — A') sin.^-^^')' 
6sin.(9 — A')sin.(ip+A')—csm.A'[s\n.( (p -{-A')-\-sin.(cp— A')]. 
But, by trigonometry, 

sin. (y -\- A') = sin. <P cos. A' ~\- cos. y sin. 4', 
sin. ( (p — A') z=- sin. <p cos .4' — cos. <p sin. ^4'; 
so that 
sin. (<p -f- -4.') + s > n - (<P — A') =2 sin. ip cos. ^4' ; 
sin. (<p — A') sin. ((p-\-A f ) = sin. 2 y cos. A' 2 — cos. 2 ^ sin. 2 A' 
= sin. 2 </>(!— sin. 2 ^4') — (1 — sin. 2 y) sin. 2 ^4' 



<§> 42.] DETERMINATE PROBLEMS. 27 

Given length intercepted. 

which, substituted in the preceding equation, give 

b sin. 2 (p — b sin. 2 A 1 = 2 c sin. <p sin. ^1' cos. A' ; 

from which we find the value of sin. ^ 

b sin. <p z= sin. A' [c cos. J.' ± \/(6 2 -|- c 2 cos. 2 A')]. 

Corollary. Of the two values of sin. ^, one is clearly nega- 
tive ; and this value corresponds to the line CB'D', which 
meets AD produced in D', so that 

ft'ty- CB' + CD'^b. 



28 ANALYTICAL GEOMETRY. [b. I. CH. III. 

Position in plane. Origin. 



CHAPTER III. 

POSITION. 

43. As almost all geometrical problems involve more 
or less the elements of Position, it is important to adopt 
some convenient method of determining and denoting 
them. We shall, at first, confine ourselves to the con- 
sideration of position in a plane, and then proceed to 
that of position in space. 

44. Problem. To determine and denote the position 
of points in a plane. 

Solution. The most natural method of determining the po- 
sition of a point is by its distance and direction ; it is thus, 
that, if a man wishes to go to any place, he starts in the 
direction of the place, and proceeds a distance equal to that 
of the place. Some point as A (fig. 17.) must then be fixed 
upon in the plane to which all the other points B, B', &>c. 
may be referred ; and the elements of position of B, B 1 , &c, 
are the distances AB, AB', &,c, and the angles which AB } 
AB', &c. make with some assumed direction, as that of AC, 
for instance. We shall denote the distances AB, AB 1 , &c. 
by r, r', &c, and the angles BAC, B'AC, &c by y, y', &>c. 

45. Definitions. The point A, which is thus fixed 
upon to determine the other points, is called the origin 
of coordinates , or simply the origin. 



§ 47.] position. 29 

Polar coordinates. Their transformation. 

The line AC is called the axis of coordinates, or sim- 
ply the axis. 

The distance of a point from the origin is called its 
radius vector, thus r, r' &c. are the radii vectores of 
B, B', &c. 

The radius vector and the angle which it makes with 
the axis are called polar coordinates. 

When the position of a point is given, its coordinates must 
be regarded as given. 

Negative radii vectores are entirely avoided by regarding 
the angles as counted from zero to four right angles. Thus 
the coordinates of B" are not the angle CAB and — AB", 
but they are AB" and — CAB', 
or 360° — CAB" — 180° + CAB' = 180° + CAB. 

46. It is often found in the course of a solution, that the 
origin and axis which have been assumed do not furnish the 
most simple results ; it is desirable, in such a case, to have 
formulae by which the elements of position can be readily re- 
ferred to some other origin and axis. 

The referring of the elements of position from one 
origin and axis to others is called the transformation of 
coordinates. 

47. Problem. To transform coordinates from one 
system of polar coordinates to another system, which 
has the same origin but a different axis. 

Solution. Let A (fig. 18.) be the origin, A C the original 
axis, and AC X the new axis. The radius vector is the same 
in both systems. Let the coordinates of any point, as B, in 
the first system, be 



30 ANALYTICAL GEOMETRY. [b. I. CH. III. 

Transformation of polar coordinates. 

AB — r, and BAC — g> j 

and let its coordinates in the new system be 

AB = r, and BAC 1 = <p x \ 
we are to find <p, in terms of y 1 . 

Now let « = CAC 1 

be the angle of the two axes, 
we have BAC = BAC X + CAC X> 

or tp — <p ! + «, 

which is the required formula for transformation. 

48. Problem. To transform coordinates from one 
system of 'polar coordinates to any other system. 

Solution. Let A (fig. 19.) be the first origin, and AC the 
axis; and let A x be the new origin, and A X C X the new axis. 
The coordinates of any point, as B, with reference to the 
first origin and axis are 

AB — r, and BAC = </> ; 
and the new coordinates are 

A X B = r lf and BA 1 C 1 = <p 1 ; 
and we are to find r and 9 in terms of r x and y x . 

The coordinates of the new origin referred to the first 
origin and axis must be known ; let them be 

AA X == «, and A x AC=z p ; 

the inclination of two axes must also be known, and let it be a. 
Produce C X A X to A', we have 

A X A'C = «, and AA'A X — 180° — «; 

also AA X A' = A X A!C — A X AA' = « — fi . 



§ 50.] POSITION. 31 

Distance of two points. 

AA X B = 180° — (BA 1 C 1 + AA X A') 

— 180° — ( Vl +a — fc) 

cos. AA X B = COS. (9j + a I s ). 

The triangle 4^1 x # gives, then, 

AB2 = A4f _f_^ ijB 2__ 2.^^. a x B. cos. ^^ X B, 
r2 = a 2 + rf + 2 a r 2 cos. (^ + a — /?) ; 
r = v /[«2_|_ r 2^_2ar 1 cos.( 9 ) 1 + a— ^)] (1) 

and J^ : A ± B = sin. (A^jB) : sin. BAA lf 

or r : r 2 ^sin.^-)-" — /?) : sin. (<p — £) ; 

whence sin. (9 — /?) = ^1. sin. (<p 1 + « — ^ (2) 

and equations (1) and (2) are the required formulae. 

49. Corollary. If the new origin is in the former axis, and 
if the axes coincide, we have 

« = 0, (5 = 0, 

and equations (1) and (2) become 

r = V(« 2 + i+2flr 1 cos. (pi ) (3) 

sin. y = — i . sin. ^ r (4) 

50. Problem. To find the distance of two given 
points from each other. 

Solution. Let B and B' (fig. 20.) be the two points whose 
coordinates are respectively r </>, and r h tp f . The triangle 
BAB' gives 

5jB / 2 _ r 2 _|__ r /2 _ 2r r' cos. (9' — 9), 
J35' = V[> 8 + r'2 _ 2 r r' cos. ( 9 ' — <p)]. (5) 



32 ANALYTICAL GEOMETRY. [b. I. CH. III. 

Distance of two points. 

51. Corollary. If the point B' is in the axis, we have 

and (5) becomes 

BB' = ^/(r 2 -L. r' 2 — 2 r r cos. cp). (6) 

52. Corollary. If 2?' is the origin, we have 

r' = 0, 
and (5) becomes 

as it should be. 

53. Corollary. If two points are upon the same radius 
vector, we have 

<P' = tp , 
and (5) becomes 

BB' &= \/(r 2 + r' 2 — 2 r r') zz= r' — r. (7) 

54. Corollary. If the two points are upon opposite radii 
vectores, we have 

<P> = 9 + 180°, 

and (5) becomes 

J3J3 = \Z(^ + r' 2 + 2 r r'J = V ^ r. (8) 

55. Although polar coordinates are the most natural ele- 
ments of position, they are not those which are usually the 
most simple in their applications. It has been found con- 
venient to adopt, in their stead, the distances from two axes 
drawn perpendicular to each other through the origin. 

The distances of a point from two axes, drawn per- 
pendicular to each other, are called rectangular co- 
ordinates. 



§ 57.] position. 33 

Rectangular coordinates. 

Thus, if XAX and YAY' (fig. 21.) are the axes, the 
rectangular coordinates of the points B, B', &c. are, respec- 
tively, BP and BR, BP' and BR, &,c. We shall denote the 
distances BR, BR 1 , &,c. from the axis YAY' by x, z', &c, 
and the distances BP, BP', &c. from the axis XAX by 
y, y 1 , &c. 

The distances x, x', &c. may be called abscissas, to distin- 
guish them from y, y', &c, which are called ordinates. 

56. When the rectangular coordinates of a point are known, 
it is easily found by measuring off its distance x from the axis 
YA Y' upon the axis XAX, and its distance y from the axis 
XAX upon the axis YAY, and the lines, which are drawn 
through the points P and R thus determined, perpendicular 
to the axes, intersect each other at the required point. 

Since the distances x, x' ', &c. are thus measured 
upon the axis XAX/, this axis is called the axis of x, 
or the axis of the abscissas ; while the axis YA Y is 
called the axis of y, or the axis of the ordinates. 

57. By using the negative sign, as in art. 4, the sign 
of the abscissa, or of the ordinate, designates upon 
which side of the axis the point is placed. 

Thus if we denote, by positive ordinates, distances 
above the axis XAX', and by positive abscissas, dis- 
tances to the right of the axis YAY', negative ordinates 
will denote distances below the axis XAX', and nega- 
tive abscissas, distances to the left of YA Y'. 

Points in the quarter YAX, being above the axis XAX 
and to the right of YA Y, will then have positive ordinates 
and abscissas. Points in the quarter YAX, being above 
XAX and to the left of YA Y, will have positive ordinates 



34 ANALYTICAL GEOMETRY. [b. I. CH. III. 

Polar transformed to rectangular coordinates. 

and negative abscissas. Points in the quarter XA Y, being 
below XAX and to the right of YAY', will have negative 
ordinates and positive abscissas. Points in the quarter X AY , 
being below XAX and to the left of YAY, will have nega- 
tive ordinates and abscissas. 

58. Corollary. For any point in the axis XAX/ the 
ordinate is zero, that is, 

2/ = 

is the algebraical condition that a point is in the axis 
of x. 

For any point in the axis YA Y 7 , the abscissa is zero, 
that is, 

x — 

is the algebraical condition that a point is in the axis 
of y. 

The coordinates of the origin are 

x = 0, y — 0. 

59. Problem. To transform from polar to rectangu- 
lar coordinates. 

Solution. Let A (fig. 22.) be the polar origin, and AC the 
polar axis. Let A 1 be the new origin, whose position is de- 
termined by the coordinates 

AA X '■=. a, A x AC—p. 

Let^LjXbe the axis of abscissas, and A X Y those of ordi- 
nates; and let the inclination of the axis i,Jto AC be a ; 
so that if the line AD is drawn parallel to A X X, we have 

« = DAC. 



§ 60.] position. 35 

Polar transformed to rectangular coordinates. 

The values of the polar coordinates 

AB — r, and BAC ' — y, 
are to be found in terms of the rectangular coordinates 

A X P = x, and BP — y. 
Produce BP to P', and A X Y to A 1 . We have 
A X AA! — A X AC—AAC = ? — «, 
BAA' = BAC — A AC = <f — «. 

The right triangles A X AA' and BAP 1 give 

A x A f = PF = ^ j. sin. A X AA' — a sin. (/?— o), 
^4 J. 7 =: ^4_4 j . cos. ^4 j.4^4 7 =± « cos. (,s — «) ; 

PP = ^i 5 + PP 1 — y + a sin. (/S — «), 
^4P — PA + .4 ,4 == x -f a cos. (/* — «) j 
^4P 2 _ (AP'y + (JBP) 2 , 

r 2 = X 2 + 2 « I' COS. (jS a) + « 2 COS. 2 (£ «), 

+ y 2 + 2 a y sin. (J — a) -f- a 2 sin. 2 (/S — a), 
— X 2 _)_ y 2 _j_2 a [xcoS.(pi— «) -fy.sin. (,* — «)]+a 2 , 
r = x /(x2 + ^ + a2 + 2a[xcos.(.5_«)+ ys in.( 1 5_a)]) ; (9) 

tang. BAP = — , 

. . y 4- a sin. ($ — a) 

tang. (,,_„) = f±__L__2; (10) 

and formulas (9) and (10) are the required formulas. 

60. Corollary. If the origins are the same, we have 
a — 0, 
and the formulas (9) and (10) become 

r = */(z 2 +y 2 ); tang, (9 — «) = ?. (11) 



36 ANALYTICAL GEOMETRY. [b. I. CH. III. 

Rectangular transformed to polar coordinates. 

61. Problem. To transform from rectangular to 
polar coordinates. 

Solution. Let AX and AY (fig. 23.) be the rectangular 
axes. Let A x be the new origin, the coordinates of which 
are 

AA' = a, and A ± A' == b. 

Let the inclination of the polar axis A^to the axis AX 
be a; so that, if A x P r is drawn parallel to AX, we have 

CA^' = a. 

The values of the rectangular coordinates 
AP 5± x, and BP — y, 
are to be found in terms of the polar coordinates 

A X B ~ r, and BA X C — y. 
We have, then, in the right triangle BA x P' t 

BA 1 P = BA 1 C + CA X P> — <p+a, 

BP' = A X B. sin. BA X P> — r sin. ( y + «), 
A^ 1 — A^.cos.BA^'zzz r cos. (y + «) ; 
whence 

x = AA' + .AP = AA' + A 1 P = a + r cos. (y + «) (12) 
y= PP-) r PB=A 1 A' + PB = b + rsm. (<*> + «) (13) 
and (12) and (13) are the required values of x and y. 

62. Corollary. If the origins are the same, we have 

a = 0, and 6 = 0, 
and the formulas (12) and (13) become 

a; = r cos. (</> + «), y — r sin. (y -|- a). (14) 



^ 64.] position. 37 



Transformation of rectangular coordinates. 



63. Corollary. If the origins are the same, and the polar 
axis coincides with the axis of x, we have 

« = 0, 
and formula (14) becomes 

x = r cos. ipj^r sin. if. (15) 

64. Problem. To transform from one system of rect- 
angular coordinates to another. 

Solution. Let AX and A Y (fig. 24.), be the axes of the 
first system ; and A X X X and A 1 Y 1 the new axes. Let the 
coordinates of the new origin A x be 

AA! = a, and A 1 A ! ±z b ; 
and let the inclination of the axis A X X X to the axis AX be 
ft, so that, if A 1 R I is drawn parallel to AX } we have 

X X A X R' = «. 
The values of the coordinates 

AP — %\ and BP — y, 
are to be found in terms of the coordinates 

A 1 P 1 =*!, and BP, —y x . 

Draw P X R parallel to AX, and P X R' parallel to AY. 
Since the sides of the angle B are respectively perpendicular 
to those of the angle P 1 A 1 R', they are equal, or 

B — a. 

The right triangles A X P X R' and BP X R give 

P X R = A X P X sin. P X A X R — x x sin. a, 
^P'^ ^Pj cos. P X A X R' — x x cos. a; 
P X R — BP X sin. P =y 1 sin. a, 

pp = pp x cos. p = yi cos > a - 



38 ANALYTICAL GEOMETRY. [b. I. CH. III. 



Transformation of rectangular coordinates. 



we also have 

A x P l —A x R l — P'R' — A x R' — P x R — x x cos.a—y x sm.a, 
BP' =z P R + BR ~ P x R-{- BR z=x x sm.*+y x cos.a; 
AP — AA> + AP — AA' + A X P', 
or x — a -\- x x cos. « — y x sin. « : (16) 

BP — PP' + BP' — A X A< + BP', 
or y z=. b -\- x x sin. « -j- y x cos. «; (17) 

and (16) and (17) are the required values of x and y. 

65. Corollary. If the origins are the same, we have 

a — 0, and 6 = 0; 
and the formulas (16) and (17) become 

x = x x cos. « — y x sin. «, (18) 

y — x x sin. « -|~ y x cos. a. (19) 

66. Corollary. If the directions of the axes are the same, 
we have 

a z=z 0, sin. BrO, COS. a =z 1 ; 

and formulas (16) and (17) become 

x= a + x x1 y = b-\-y x . (20) 

If the new origin is, in this case, in the axis AX, we have 

b — 0; 
and formulas (20) become 

x*=a-\-z lt y — y x . (21) 

But if the new origin is in the axis AY, we have 

a — 0, 
and formulas (20) become 

x = x 19 y = 6 +y v (22) 



§ 70.] position. 39 

Distance of two points. Oblique axes. 

67. Problem. To express the distance between two 
points in terms of rectangular coordinates. 

Solution. The required result might be obtained by substi- 
tuting in formula (5) the values of r, r', y, and <p' obtained 
from formula (11), by taking x' and y' to correspond to r' and 
<p'. But it is more readily obtained by direct investigation. 

Let the two points be B and P' (fig. 25.), whose coordi- 
nates are respectively x, y, and %', y'. Draw BR parallel to 
AX, and the right triangle BB 1 R gives 

BB 2 — BR? + BR 2 = (x< — xf + (y 1 — yf 
BB' = */[(* - xf + (ij - yf]. (23) 

68. Corollary. If one of the points B is the origin, we 
have 

x = 0, y = 0, 
whence 

AB' — */(x'2 + y'*). (24) 

69. Instead of the two axes being at right angles ^o each 
other, they are sometimes taken at any angle whatever ; and 
instead of the distances from the axes, the lengths of lines 
drawn parallel to the axes are used. 

In this case, the axes and coordinates are said to be 
oblique. 

Thus if the axes are AX and A Y (fig. 26.), the coordinates 
of B, B', &c, are, respectively, 

x =z AP — BR, and y = BP = AR, 
and 

x' = AP 1 — BR 1 , and y = .B'P' = AR', &c. 

70. Problem. To transform from one system of ob- 
lique coordinates to another. 



40 ANALYTICAL GEOMETRY. [b. I. CH. III. 

Transformation of oblique coordinates. 

Solution. Let ^IXand ^4F(fig. 27.) be the original axes, 
their mutual inclination being 

Let the new axes be A X X X and A X Y X , which are inclined to 
the axis AX by the angles « and /*, so that, if A X P' is drawn 
parallel to AX, we have 

X X A X P = a, and Y X A X P — P. 
Let the coordinates of A 1 be 

A A' = a, and A X A' = 6. 
The values of the coordinates 

AP ±z z, and BP — y, 
are to be found in terms of 

A X P X =zx lt and BP, = y x . 

Draw P X R parallel to AY, and P X R parallel to AX. We 

have, then, 

P X RP — P X R'P> = YAX— y, 
A X P X R — P X RP — P X A X R = / — «; 
BP X R' — Y X A X R ac /?, 
B — P X RP' — BP X R' = Y — P; 
sin. A X RP X : sin. A X P X R — A X P X : A X R, 

or sin. y : sin. (y — <*) — x. : A X R, 



and 



Ai r^;^^±-1^ ; 



sin. y 



sin. A X RP X : sin. P X A X R r= vi^ : jP^ 

or sin. y : sin. a =z x x : P X R, 

and Pl R = P>Il' = *^;. 

1 Sill, y 



§ 72.] position. 41 

Transformation of oblique coordinates. 

sin. BR P 1 : sin. B = BP \ : P,R', 

or sin. y : sin. (y — («) = y x : P X R', 

and P 1 R' = KP' = ^ S ' m /^--Il; 

sin. 7 

sin. BRP 1 : sin. ^P^' = BP X : .BiZ , 

or sin. y : sin. f* •== yj : jBi?', 

and W^ftfo* 

sin. y 

AP=AA-\-AP~AA+A l P=AA+A 1 R + RP ) 

. x sin. (•/ — «) + y, 'sin. (y — ,«) ,_„,. 

or a- = a H — ! 5 7 , ' - 71 ^ -' ; (25) 

sin. y 

#P — PP' + P'B = A ± A' + PR' + RB, 

x sin. « -f- y sin. ? 

or ?/ — &-] . ' ^ : (26) 

J ' sin. y v 7 

and (25) and (26) are the required values of x and y. 

71. Corollary. If the original axes are rectangular, we have 

y = 90°, sin. y=l, 
and formulas (25) and (26) become 

x z=z a -\- x x cos. « -(- y x cos. /?, (27) 

y — b -\- x x sin. « + y x sin. (?: (28) 

72. Corollary. If the new axes are rectangular, we have 

j* = 90° -\- «, sin. |S =— cos.' «, 
sin. ( y — (?) = sin. (y — « — 90°) == — cos. ( y — «) j 
and formulas (25) and (26) become 

, _ a + ;.' si "-^— Q-y,coe.(r-^> (29) 

1 sin. y 

7 sin. y 

4* 



42 ANALYTICAL GEOMETRY. [fi. I. CH. III. 

Point in space. Rectangular axes. 

73. Problem. To determine the position of points in 
space. 

Solution. The most natural method of determining the 
position of a point in space is to determine the position of 
some plane passing through the point, and then to determine 
the position of the point in this plane. For this purpose some 
fixed axis AC (fig. 28.) is assamed, and some fixed plane CAD 
passing through this axis. The plane CAE, which passes 
through the point B and the axis AC, is determined by the 
angle EAD, which it makes with the fixed plane CAD. 
The position of the point B in the plane CAB is determined 
by the radius vector AB and the angle BAC, which this 
radius makes with the axis. The same method may be adopt- 
ed for any other points B' , B", &c, which are not given on 
the figure, as they would only render it confused. We may, 
then, denote these radii vectores of the points B, B' , &c. by 
r, r' t &lc. ; the angles which these radii make with the axis 
AC by (p, <p' t &c. ; and the angles which the planes in which 
they are contained make with the fixed plane by 6, 6' f &,c. 

74. A system of rectangular coordinates in space has 
been adopted similar to those in ] plane, and possessing 
the same practical advantages of simplicity. 

For this purpose three planes XAY, YAZ, and XAZ 
(fig. 29.) are drawn perpendicular to each other, and the rect- 
angular coordinates of a point are its distances from these 
planes. Thus, if the point B is taken, and the perpendiculars 
BP, BQ, and BR are drawn perpendicular to the given 
planes, these distances are the rectangular coordinates of B. 
If these coordinates are given, the point B is determined, by 
taking 

AL = BR, AM— BQ, and AN = BP, 



§ 76.] position. 43 

Projection of a point. 

and drawing planes through the points L, 31, and JV parallel 
respectively to the given planes, and their intersection B is 
the required point. 

The intersections AX, AY, and AZ of the three 
given planes are called the axes. 

If the coordinates of the point B are 

x = AL = BR = NQ = MP, 
y = AM = BQ = LP = NR, 

z = AN = BP = 31R = LQ, 

the axis AX is called the axis of x, AY is called the axis of 
y, and AZ the axis of z ; the plane XA Y is called the plane 
of xy, the plane XAZ is called the plane xz, and the 
plane YAZ is called the plane of y z. The coordinates of 
B', B", &c. are, in the same way, 

x 1 = AL', y' = B'Q', z< = L'.Q', 
and 

x"= AL', y" = B"Q", z" = R"Q" S &c. &c. 

75. The foot of the perpendicular let fall from a 
point upon a plane is called the projection of the point 
vpon the plane. 

Thus the projection of B upon the plane of x y is P the 
coordinates of which are x and y ; the projection upon the 
plane of xz is Q, the coordinates of which are x and z * the 
projection upon the plane of y z is R, the coordinates of which 
are y and z. 

76. Problem. To transform from, rectangular co- 
ordinates to the polar coordinates of art. 73, the origins 
being the same, the polar axis being the axis of x, and 
the fixed plane the plane of x y. 



4 «0 | 






44 ANALYTICAL GEOMETRY. [b.I. CH. III. 

Polar coordinates transformed to rectangular. 

Solution. Let AX, AY, AZ (fig. 30.) be the rectangular 
axes. Let XAD be a plane passing through the point J9y_ 

The values of 

AL = x, BQ = y, and CQ = z, 
are to be found in terms of 

AB = r, BAX— < P> and DAY=z 6. 

Join BL. In the right triangles ABL or BLQ, we have 
the angle 

because the sides LB and BQ are parallel to AD and _4 Y; 
we also have 

x — AL = AB cos. J5yli == r cos. </>, (31) 

^X = ^41? sin. 2L1L == r sin. </> ; 

y = J3Q x= -B£ cos. LBQ = r sin. y cos. a, (32) 

z = XQ == BL sin. ZjBQ == r sin. <p sin. 0. (33) 

77. Problem. To transform from the polar coordi- 
nates of art. 73, to rectangular coordinates, the origins 
being the same, the polar axis being the axis of x, and 
the fixed plane the plane ofxy. 

Using the figure and notation of- the preceding article, the 
values of r, g>, and &, are to be found in terms of x, y, and z. 
They may be immediately found from equations (31), (32), 
and (33). The sum of the squares of these equations is 
x 2 _|_ «2 _l_ %% = r 2 (cos. 2 c/> -(- sin. 2 y cos. 2 (3 + sin. 2 y sin. 2 d) 
== r 2 [cos. 2 ^ + sin. 2 y (cos. 2 6 + sin. 2 a)] 

= r 2 (cos. 2 9 + sin. 2 y) == r 2 ; 
because, 

1 — cos 2 6 -(" sn1 - 2 3= * cos - 2 y ~\~ s ^ n * 2 V 



§ /y.J POSITION. 


45 


Distance of two points in space. 


Hence 




r = </(* 2 + y 2 + *), 


(34) 


and (31) gives 




X X 
COS. 9= = ,^ : 3 : _ ox . 


(35) 



The quotient of (33) divided by (32) is 

sin. 6 z . _ 

= tang, ft = -. (oo) 

cos.d y 

78. Problem, To find the distance apart of two 
points. 

Solution. Let 5, P' (fig. 31.) be the points, and P, P' their 
projections upon the plane of xy. Join PP' , and draw J5£ 
parallel to PP . Since p, P' are two points in the plane YAX, 
we have, by equation (23), 

pp /2 = (i / — *) 2 + (</' — y-) 2 J 

and in the right triangle BEB 1 , 

BE — B'P' — BP — z—z, 
BB 2 = PJE? 2 + BE 2 = PP 2 + JB'.E 3 , 

= (z'-z) 2 + (y-y) 2 + (z'-*) 2 ; 

PP' = v[(*'-*) 2 + (y--y) a + (* f -*) 2 ]- ( 3 ~) 

79. Corollary. If one of the points, as P 7 , is the origin, we 
have 

x 1 = 0, y' = 0, z> == 0, 

and formula (37) becomes 

4J3 = V(x2 + y 2 + * 2 ), (38) 

which agrees with equation (34). 



46 ANALYTICAL GEOMETRY. [b. I. CH.1II. 

Projection of a line. 

80. The line PP' : which joins the projections of the 
Jwo extremities B, B' of the line BB', is called the 
projection of the line BB' upon the plane of xy. 

81. Corollary. If the angle B'BE, which is the inclina- 
tion of the line BB' to its projection or to the plane of y x, is 
denoted by x, the right triangle B'BE gives 

BE = BB' cos. B'BE, 
or 

PP' — BB 1 cos. X; 

that is, the projection of a straight line upon a plane is equal 
to the product of the line multiplied by the cosine of its in- 
clination to the plane. 

82. If planes BPL, B'P'L', are drawn through the 
extremities B, B' of a line BB' , perpendicular to an 
axis AX, the part Eh' intercepted between these lines 
is called the projection of the line upon this axis. 

83. Corollary. Since 

LL' =z AL' — ALz=x' — x, 

the projection of a line upon an axis is equal to the 
difference of the corresponding ordinates of its extremi- 
ties. 

The projection of the radius vector AB is AE, or 
the corresponding coordinate of its extremity. 

84. Corollary. It follows from equation (37), that 
the square of a line is equal to the sum of the squares 
of its projections upon the three rectangular axes. 



<§> 87.] POSITION. 



Sum of the squares of the angles made by a line with the axes. 



85. Corollary. If the inclination of the line AB to the 
axis AX is denoted by ip, we have, by drawing LS parallel 
to AB to meet the plane B'P'L' in S, 

LS = BB 1 , SLL' = j, 
LL< = LS cos. SLL', 
LL' = BB' cos. (p ; 

that is, the projection of a straight line upon an axis 
is equal to the product of the line multiplied by the 
cosine of its inclination to the axis. 

86. Corollary. If v is the inclination of the line to the axis 
of y, and to its inclination to the axis of z, its projections upon 
the axes are, respectively, 

BB' cos. y, and BB' cos. a> ; 

so that, by art. 82, 

BB® = BB® cos. 2 cp+BB'2 cos.2 y + ££' 2 cos. 2 w , 

or, dividing by ^5 /2 , 

1 = COS. 2 V -{" COS. 2 V + COS 2 w ; 

that is, fAe sm??& o/7Ae squares of the angles which a 
line makes with three rectangular axes is equal to 
unity . 

87. A different system of polar coordinates from that 
of art. 73, is often used upon account of its symmetri- 
cal character. It consists in determining the direction 
of the radius vector by the angle which it makes with 
three rectangular axes. 



48 ANALYTICAL GEOMETRY. [B. I. CH.-III. 



Rectangular transformed to oblique coordinates. 



88. Corollary. If the angles tp, v, and « denote the angles 
which the radius vector makes with the axes of x, y, and z, 
we have, by arts. 83 and 84, 

x — r cos. <p, y z= r cos. i//, z =± r cos. to ; (39) 

cos. 2 g> -]- cos. 2 v -f- cos. 2 w = 1, 

which will serve to transform from rectanglar to polar 
coordinates in the system of the preceding article. 

89. Oblique coordinates are sometimes used similar to 
oblique coordinates in a plane ; thus, if the axes AX, A Y, 
and AZ (fig. 29.) had been oblique to each other, and the 
other lines drawn parallel to the axes, the point B would be 
determined by the oblique coordinates 

AL = x, LQ = y, QB = z ■; 

and, in the same way, for other points B', B", &c. 

90. Problem. To transform from rectangular to ob- 
lique coordinates. 

Let AX, AY, AZ, (fig. 32.) be the rectangular axes, and 
A 1 X 1 , A 1 Y 1 , A 1 Z 1 , the oblique axes. Let the coordi- 
nates of the new origin be 

AA 1 = a, A 1 A" — b, A 1 A" == c ; 

and let the inclination to the axes AX, AY, and AZ of those 
axes A X X X be, respectively, «, a', a"', let those of the axis 
A 1 Y 1 be |S, p, ?"; and those of the axis A 1 Z 1 be y, /, /'; 
these angles must be subject to the condition of art. 84 ; 
that is, 

COS. 2 a -|- COS. 2 "' -|- COS. 2 <*" = 1, 
COS. 2 ,* -f- COS. 2 /*' -f- COS. 2 '?' = 1, 

cos. 2 y -j- cos. 2 y' -(- cos. 2 y" ztz 1. 



§ 91.] position. 49 

Oblique transformed to rectangular coordinates. 

The values of the rectangular coordinates 

AL = x, BQ = y, LQ = z, 
are to be found in terms of 

A X L X =x L , BQ 1 = y x , L x Q t =z 1 . 

Let L' and Q be the projections of the points L x and Q x 
upon the axis of z. Since A'L' t L'Q\ and Q'L are the 
projections, respectively, of A x L lt L 1 Q 1 , and Q X B, upon 
the axis of x ; and A r L ±) Q 1 B, and L 1 Q 1 , are respectively 
parallel to the axes of x lt y lt and z. f and, therefore, inclined 
to the axis of x by the angles *, (f 3 and v, we have 

A'L' = A 1 Ii 1 COS. a = Xj COS. or, 

Q'L = Q X B cos. (? = ajf, cos. ft 

£'Q' = ijQj cos. ■/ = Zj cos. y ; 
so that 

^i =: AA' + 4'Z' + Q'L + ZQ' 

gives 

x = a -\- z x cos. a -\- y x cos. /s -[~ *i cos - *■ (40) 

In the same way we might find 

y = 6 -f- *i cos - « H~ #i cos - |S ' + % i cos - "/• (^1 ; 
2=c-f x i cos - «" + #i cos - •*" + z i cos - y// 5 (4-) 

so that equations (40), (41), and (42) are the required equa- 
tions. 

91. Corollary. If the new axes are also rectangular, equa- 
tions (40), (41), and (42) may still be used, but the angles 
«» /*» Y > a 'y ?'> y't a "> ?"> an( ^ Y " W *M be subject to certain con- 
ditions, which are thus obtained. Let r be the radius vector 
drawn from A x to B, and let the angles which r makes with 

5 



50 ANALYTICAL GEOMETRY. [b. I. CH.1II. 

Angle of two lines. 

the axes AX, AY, AZ, A 1 X 1 , A X Y X , A 1 Z X , be re- 
spectively (p, yj, o), v lt y lt W j, we shall have, by art. 83, 

x — AL = AA' -\- A'L = a -f- r cos. ^, 
x x — r cos. y 1 , y x = r cos. V x , z x — r cos. Wj ; 

which may be substituted in equation (40). If, in the result, 
we suppress the common term a, and the common factor r, we 
have 

COS. <p = COS. a cos. (p x -)- cos. p COS. v j + cos. y cos. a, x ; (43) 

which expresses the angle cp made by two lines, one of 
which is inclined to the three axes of x x , y x , z x , by the 
angles «, /*, y ; and the other by the angles <fi, Vi> »i« 

This formula may then be used for determining the angle 
which any two lines make with each other, and which are 
inclined to the axes of z, y, z by given angles ; to determining 
the angles, for instance, which the axes of x x , y x , z 1 make 
with each other. But these axes are perpendicular to each 
other, and therefore we have for the angles of x x and y x , of 
Xj and Zj, of y x and z x , respectively, 

cos. 90° == = cos. « cos. /? -|- cos. a 1 cos. p? -\- cos. a" cos. p' 1 , ( . 4) 
cos. 90° — =. cos. a cos. y -\- COS. «' COS. y'-[~ cos. «" COS. Y l! , (45) 
COS. 90° = = COS. (5 COS. Y + COS. ?' COS. y' + cos. |*" cos. y". (46) 

92. Corollary. By applying the preceding formulas to the 
axes of x, y, z, referred to those of x x , y x , z x , we have 

cos. 2 « + cos. 2 i 5 + cos. 2 y — 1, (47) 

cos . 2 a' -j_ COS. 2 F + COS. 2 y' = 1, (48) 

cos. 2 a" -f- cos. 2 r _[_ cos. 2 y" = 1 ; (49) 



<§> 93.] position. 51 

Change of origin. 

COS. a COS. a' -f- COS. p COS. p' -f- COS. y COS. y' = 0, (50) 

COS a COS. a" -\- COS. p COS. p" -f- COS. y COS. y" z= 0, (51 ) 

COS. a ' COS. «" -|- cos. p' COS. ( «" -}- COS. y' COS. y" = 0. (52) 

93. Corollary. If the origin is changed, but not the direc- 
tions of the axes, we have 

a — 0, p — 90°, y = 90°, 

a — 90°, /J' =0, y' — 90°, 

«"z=90°, P"— 90°, y "= 0; 
and equations (40, 41, 42) become 

x—a-\-x x (53) 

#=& + #! (54) 

*=* + *,. (55) 



£2 ANALYTICAL GEOMETRY. [b. I. CH. IV. 

Loci. Angles. 



CHAPTER IV. 

EQUATIONS OF LOCI. 

94. When a geometrical question regarding position 
leads to a number of equations less than that of the 
unknown quantities, it is indeterminate, and usually 
admits of many solutions • that is, there are usually a 
series of points which solve it, and this series of points 
is called the locus of the question, or of the equations to 
which it leads. 

95. The equation of the locus of a geometrical ques- 
tion is found by referring the positions of its points to 
coordinates, as in the preceding chapter, and expressing 
algebraically the conditions of the question. 

96. Scholium. Instead of denoting angles by degrees, 
minutes, &c. ; we shall hereafter denote them by the 
lengths of the arcs which measure them upon the cir- 
cumference of a circle whose radius is unity, and shall 
denote by n the semicircumference of this circle, which 
is nearly 3-1415926. 

The angle^of 90°, or the right angle, is thus denoted by 
J n, the anglfc of 180°, or two right angles, by tt, and the 
angle of 360°, or four right angles, by 2 ™ t 

97. Corollary. The arc which measures an angle 6 in the 






*§> 98.] EQUATIONS OF LOCI. 53 

Circle. 

circle whose radius is R is R 6, because similar arcs are pro- 
portional to their radii, and 6 *is the length if the radius is 
unity. 

98. Examples. 

1. Find the equation of the locus of all the points 
in a plane, which are at a given distance from a given 
point in that plane. This locus is the circumference of 
the circle. 

Solution. Let the given point A (fig. 33.) be assumed as 
the origin of coordinates, and let R = the given distance. 

If the polar coordinates of art. 44 are used, we have for 
each of the required points, as M } 

r = R; (56) 

so that this equation is that of the required locus. 

Corollary 1. Equation (56) is the polar equation of 
the circle whose radius is R, and centre at the origin. 

Corollary % Equation (56) may be referred to other polar 
axes by arts. 48 and 49. Thus for the point A xt for instance, 
for which 

AA X ■==. a = — R „ 

equation (3) becomes 

r = x/(R* +r\—ZRr 1 cos. 9l ) 
which substituted in (56) gives, by squaring and reducing, 
r\ — 2 Rr x cos. (p x = ; 

or we may divide by r lt since r x is not generally equal to 
zero, and the equation is 

r x —2Rcos. y x (57) 

5* 



54 ANALYTICAL GEOMETRY. [b. I. CH. IV. 

Circle. 

which is the polar equation of a circle whose radius is 
R, the origin being upon the circumference, and the 
line drawn to the centre being the axis. 

Corollary 3. Equation (56) may, by art. 60, be referred to 
rectangular coordinates; and equations (11) being substituted 
in (56), and the result being squared, we have 

X 2 -f y2 ~ R2 (58) 

which is the equation of a circle whose radius is R, 
referred to rectangular coordinates, the origin of which 
is the centre. 

Corollary 4. Equation (58) may, by art. 66, be referred to 
any rectangular coordinates. Thus the axes A 2 X 2 , A 2 F 2 , 
for which the coordinates of A are A 2 A 1 and AA' t so that 

a = — A 2 A' = — a', 

b = — A A' = — b', 
give 

x = x 2 — a>, y = y 2 — b ! , 

which, substituted in (58), give 

(*, - a V + to - *r = R2 < (59) 

which is the equation of a circle, referred to rectangular 
coordinates, the radius of the circle being R, and the 
coordinates of the centre of and b'. 

Corollary 5. For the point A t we have 
a' = R, b' = 0, 
so that for this point (59) becomes 

(i,— B)«+y» = R», 
or x\ — 2Rz t + R 2 -\-y\ = R* 

y2=2Rz,—x\, (60) 



1) & •+ 






§ 98.] EQUATIONS OF LOCI. 55 



Sphere. 



which is the equation of a circle, whose radius is R, 
referred to rectangular coordinates, the origin of which 
is upon the circumference, and the axis of x 1 is the 
diameter. 

2. Find the equation of the locus of all the points in 
space, which are at a given distance from a given point. 
This locus is the surface of the sphere. 

Solution. Let the given point be assumed as the origin of 
coordinates, and let 

R = the given distance. 

If polar coordinates are used, we have for each of the re- 
quired points 

r = R, (61) 

which is, therefore, the polar equation of a sphere, 
whose radius is R, and centre at the origin. 

Corollary 1. Equation (61) may be referred to rectangular 
coordinates, by art. 77 ; and if equation (34) is substituted in 
(61), and the result squared, we have 

X 2 _|_ y 2 _j_ Z 2 = R2 . (62) 

which is the equation of the sphere whose radius is 
R, referred to rectangular coordinates, the origin of 
which is the centre. 

Corollary 2. Equation (62) may, by art. 93, be referred to 
any rectangular coordinates, and the substitution of equations 
(53, 54, 55) in (62) gives 

(*i + <0 2 + &, J r &) 2 + (*, + c? = * 2 > (63) 
v)hich is the equation of a sphere referred to rectangular 



56 ANALYTICAL GEOMETRY. [b. I. CH. IV. 



Ell] 



pse. 



coordinates, the radius of the sphere being R, and the 
coordinates of the centre — a, — b, and — c. 

3. Find the equation of the locus of all the points 
in a plane, of which the sum of the distances of each 
point from two given points in that plane is equal to a 
given line. This locus is called the ellipse, and the 
given points are called its foci. 

Solution. Let F and F (fig. 34.) be the foci, let F be the 
polar origin, let the line FF joining the foci be the polar 
axis, and let 

2c = FF' === distance between the foci, 

2 A = the given length ; 

where the length A is not to be confounded with the point A 
of the figure. 

If, then, we put in equation (6) 

r' = FF' = 2e, 

we have for the distance MF' of each point M from F' t 

MF' = \/(r 2 -f 4 c 2 — 4 c r cos. y) ; 
so that 

FM+ MF' = 2 A =r +V(r2-)-4c 2 — 4crcos.y) 
\/(r 2 -f- 4 c 2 — 4tcr cos. y) = 2 ^4 — r, 
and squaring and reducing 

4c 2 — 4 c r cos. fp = 4 A 2 — 4ir 
(-4. — ; c cos. </>) r = A 2 — c 2 

r = 4 — . (64) 

^4 — c cos. /n 



<§> 98.] EQUATIONS OF LOCI. 57 



Ellipse. Transverse axis. 

which is the polar equation of the ellipse, one of the foci 
being the origin, and the axis being the line joining 
the foci, which is called the transverse axis, if it is pro- 
duced to meet the curve. 

Corollary 1. For the point C where the ellipse cuts the 
axis, we have 

cp =z 0, COS. (f z=z 1, 
42 c 2 

FC—r= — A =A + c. 

A — c ' 

Corollary 2. For the point C, where the ellipse cuts the 
axis produced, we have 

(9 = TV, COS. (p =z l r 

A 2 f 2 

FC=r = A A , = A — c. 
A -\- c 

CC = FC+FC = 2 A; 

so that the transverse axis is equal to the sum of the 
distances of each point of the ellipse from the two foci. 

Corollary 3. If FF' is bisected at A, we have 
AF — AF' — c, 
AC— FC—AF=zA + c—c = Az=iCC'=iAC / ; 

A is called the centre of the ellipse. 

Corollary 4. If we put 

B 2 = A 2 — c 2 , 
(64) becomes 

B 2 

r = — - . (65) 

A — c cos. 9 



58 ANALYTICAL GEOMETRY. [b. I. CH. IV. 



Ellipse. Eccentricity. 

Corollary 5. If we put 

e __ 2c 

e is called the eccentricity of the ellipse, and is the ratio 
of the distance between the foci divided by the transverse 
axis. 

Hence c — A e, 

and this, substituted in (64), gives 

. = A» (!-« ») = A(l-e*) _ 
A (I — e cos. </>) 1 — e cos. y 

If we also put 

FC — P ~A — c — 4(1 — e), 

(66) may be put in the form 

■ A{l—e)(l+e) P{l+e) 



1 — e cos. ip 1 — ecos. 



(6?) 



Corollary 6. The equation of the ellipse may be referred 
to rectangular coordinates by arts. 59 and 60. Thus, if we 
take the point A for the origin, and AC for the axis of x, we 
have 

a — : FA — c, 

a — ? — 0, sin. ( p — «) = 0, cos. (/? — «) = 1 . 

whence (9) and (10) become 

r = x/(* 2 + 3/ 2 + c 2 + 2 c x) , 



§ 98.] EQUATIONS OF LOCI. 59 



Ellipse. 


Conjugate axis. 


and 

1 


1 x + e 


sec. q> 


~ V(l + tan. 2 y) ~ V(^ 2 + 2cx+c 2 +x 2 ) 

x 4- c 
cos. (p — — ■ — : 



r 
r cos. <p — x -f- c. 
Now equation (64) freed from fractions is 
A r — cr cos. 9 = A 2 — c 2 ; 

in which, if we substitute the preceding values, we have 

As/{x 2 -i r y2-i rC 2 _|_ 2 c z) — c s — c 2 —A 2 — c 2 , 

or 

A\/{x 2 + y 2 + c 2 + 2 c x) — A 2 + c x. 

The square of which is, when reduced, 

A 2 x 2 + A 2 y 2 + ^ 2 c 2 = 4* _|_ C 2 ^V 
or 

(^2 _ C 2) a;2 ^ ^2 y 2 _ ^4 _ ^2 c 2 — £2 (^2 _ c 2^ 

or substituting _B 2 

£ 2 x 2 + ^4 2 y 2 — 4 2 jB 2 ; (68) 

which, divided by A 2 B 2 , is 

x 2 II 2 

A*+h= h < 69 > 

Corollary 7. The part 5 B' of the axis of y included 
within the ellipse is called its conjugate axis. 

We have for the points B and B', art. 58, 

x — 0, y = AB or = — 4.B' ; 






60 ANALYTICAL GEOMETRY. [b. I. CH. IV. 

Ellipse. Conjugate axes. 

which, substituted in (6S), gives 
A 2 y' 2 = A 2 B 2 , 

y' = d= B — + jBor = — .B; 
so that ^4J5 == ^4.8' = B, 

and 5 «s equal to the semi conjugate axis. 

Corollary 8. Equations. (68) and (69) are the equa- 
tions of an ellipse referred to rectangular coordinates, 
the centre of the ellipse being the origin, the transverse 
axis 2 A being the axis ofx, and the conjugate axis 2 B 
being the axis of y. 

Corollary 9. The equation of the ellipse may, by art. 71, 

be referred to oblique axes. Thus, if the origin is unchanged, 

we have 

a = 0, 6 = 0, 

and equations (27) and (28) become 

X = X x COS. a -}- y ^ COS.-/?, 

y = % sin. a -\- y ± sin. p; 
which, substituted in (G8), give, by simple reduction, 
(B 2 cos. 2 «+ A 2 sin. 2 «)x f-j-2( B 2 cos.« cos.j?-f A 2 sin.« sin.?) z y y , 
+ (£ 2 cos. 2 1? + A 2 sin. 2 (?) y J = A 2 B 2 (70) 

Corollary 10. If " and /», instead of being taken arbitra- 
rily, are so taken that we have 

B 2 cos. « cos. /? + A 2 sin. a sin. £ = 0, (71 ) 

or A 2 sin. « sin. P = — B 2 cos. « cos. /?, 

or, dividing by A 2 cos. a cos. /?, 

S 2 

tang. « tang. /? = — — , (72) 

the axes are said to be conjugate to each other. 



$ 98.] EQUATIONS OP LOCI. 61 

Conjugate diameter. 

Equation (70) is then reduced to 
(JB 2 cos. 9 «+JL 2 sin. 2 «)x2-f(JB2cos.^+^ 2 sin.^)yf_^2j B 2. (73) 

Corollary 11. If C 1 A C\ (fig. 35.) is the axis of x^ and 
B X AB\ is the axis of y x , we have for the points C x and C' x , 
where the axis of x x meets the curve 

y x ~ 0, x\ — AC X or — AC X ; 

which, substituted in (73), give 

(B 2 cos. 2 « -f A* sin. 2 «) *;§ — ^2^2 

AB 



«i==db 



\/(,B 2 cos. 2 a -\-A 2 sin. 2 «) 5 



so that the distances AC x and AC\ are equal, and in 
general any line, which passes through the centre and 
terminates in the curve, is bisected, and is hence called 
a diameter. The axes of x x and y x , which are sub- 
ject to the condition of equation (71) or 72), are called 
conjugate diameters, and equation (73) is the equation 
of the ellipse referred to conjugate diameters, which are 
inclined by the angles a and P to the transverse axis. 

Corollary 12. If we take 

B —AB, == AB 



we have 



A 1 — 



B = 



i ' 
AB 



V(^ 2 COS. 2 «-j-^2 smi 2 a )> 

AB t 

V(£ 2 cos. 2 p + A 2 sin. 2 p) ' 



o2 ANALYTICAL GEOMETRY. [b. I. CH. IV. 

Hyperbola. 



so that 



AB 

\/(B 2 cos. 2 « + A 2 sin. 2 «) = — - 



\/{B 2 cos. 2 p + A 2 sin. 2 /?) = 



.4B 



which, substituted in (73), and the result divided by ^4 2 jB 2 , 
give 



A' 2 ' # 2 
or 

JB 2 z 2 + ^1 /2 «/ 2 = A' 2 B 2 ; (75) 

which are, therefore, precisely similar in form to equa- 
tions (68) and (69) ; and they are the equations of the 
ellipse referred to the conjugate diameters 2 A' and 
%B'. 

4 Find the equation of the locus of all the points in 
a plane, of which the difference of the distances of each 
point from two given points in that plane is equal to a 
given line. This locus is called the hyperbola, and the 
given points are called its foci. 

Solution. Let F and F (fig. 36.) be the foci, let F be the 
polar origin, let the line FF' joining the foci be the polar 
axis, and let 

2c ~ FF' = the distance between the foci, 
2 A = the given length. 
If we put, in equation 6, 

r' — FF' = 2 c, 
w,e have, for the distance of each point M from F\ 
MF 1 = \/{r 2 -f 4 c 2 — 4 c r cos. <p) ; 



"§> 98.] EQUATIONS OF LOCI. 63 

Polar equation of hyperbola. 

so that 

MF — MF' = 2 A — r — x /(r2-j-4c 2 — 4 cr cos. y), 

\/( r 2 -|- 4 e 2 — 4 c r cos. (?) —r — % A. 

Squaring and reducing, we have 

4 c 2 — 4 c r cos. </> = — 4ir-|-4 ^4 2 

(A — c cos. 9) r = ^4 2 — c 2 

A 2 — c 2 c 2 — A 2 

r=- A = -\ (76) 

A — c cos. ( p c cos, cp — A v ' 

which is the polar equation of the hyperbola, one of the 
foci being the origin, and the axis being the line xohich 
joins the foci, the part of which CO ', intersected by the 
curve, is called the transverse axis. 

Corollary 1. If equation (76) is compared with (64), 
it appears that these equations have the same form, and 
only differ in the circumstance, that c is Jess than A for 
the ellipse, and greater than A for the hyperbola. In 
the ellipse, then, the value of r is always positive, be- 
cause tne numerator A* — c 2 is positive, and so is the 
denominator A — c cos. cp. For c cos. y is less than c, 
and therefore less than A. But in the hyperbola, while 
the numerator c 2 — A? is positive, the denominator 
c cos. cp — A is only positive when 

c cos. cp > A, 

A 

or cos. cp > — . 

c 

A 

If then we take cos. cp = — , 9 must be confined to 

the limits <p and — cpo. 



64 ANALYTICAL GEOMETRY. [b. I. CH.IV. 

Polar equation of hyperbola. 

Corollary 2. The above solution is limited to the 
hypothesis, that FM is greater than F' M, but if it 
were supposed less, we should have the equation of 
another curve situated with reference to the foci F' 
and F precisely as the curve of equation (76) is with 
regard to the foci F and F' . 

Since both these curves satisfy the conditions of the 
problem, they are included in the common name of 
an hyperbola, and are called its branches. 

To find the equation of the second branch referred to the 
same polar coordinates as those already used, we have 

F'M 1 — FM 1 = 2A = V(r 2 + 4 c 2 — - 4 cr cos. 9>) — r, 
\/(r 2 -f- 4 c 2 — 4 c r cos. <p) — 2A-\-r. 
Squaring and reducing, we have 

c 2 — cr cos. cp = A 2 -f- A r, 
(A -f- c cos. y)r = c 2 — A 2 , 

* 2 —A 2 = 

r = IT i » ( 77 ) 

A + c cos. <p v ' 

which is the polar equation of the branch C M lt the 
focus F being the origin, and the line joining the foci 
being the axis. 

Corollary 3. The numerator of equation (77) is posi- 
tive, but the denominator is negative, when 

A + c cos. <p< 0, 

c 
or cos. 9 < j, 



or 



cos. <p< — cos. 9 ; 



§ 98.] EQUATIONS OF LOCI. 65 

Transverse axis of hyperbola. Centre. 

or when 9 is included between the limits n — g> and 
n + ^oj so tnat 9 mus t not be taken between these 
limits. 

Corolla?-?/ 4. For the point C, where the first branch of the 
hyperbola cuts the axis, we have, by equation (76), 

cp = 0, COS. (p zzz 1, 

c 2 A2 

FC=r= - c + A . 



Corollary 5. For the point C, where the second branch 
cuts the axis, we have, by (77), 

<P = 0, COS. (p z=z 1, 

c 2 a 2 

FC'=r = , " = c — A. 
c + A 

Hence CC — FC — FC = 2 A ; 

and the transverse axis is equal to the difference of the 
distances of each point of the hyperbola from the two 
foci. 

Corollary 6. If FF' is bisected at A, we have 

AF — AF> = c, 
AC — AF—AC = A + c — c = A = JCC = .4C. 
J. is called the centre of the hyperbola. 

Corollary 7. If we put 

B 2 = c* — ^ 



66 ANALYTICAL GEOMETRY. [b.I. CH. IV, 

Conjugate axis of hyperbola. Eccentricity. 



(76) and (77) become 



c cos. cp — A 

A-\- C COS. y 



(78) 
(79) 



If this value of B 2 is compared with that of the ellipse 
of corollary 4. we see that it is, in form, the negative 
of it. 

2 B is called the conjugate axis of the hyperbola, and 
is laid off upon the line BAB' drawn through the cen- 
tre perpendicular to the transverse axis, taking 

AB = AB' — B. 

Corollary 8. If we put 

c 2c 1 

e = -A = 2ri = -^v > = sec '^ 

e is called the eccentricity of the hyperbola. 

We have c — e A, 

and this, substituted in equations (76) and (77), gives 

■*(>-«») ^("-i) (80) 

1 — e cos. (p e cos. y — 1 



A(e 2 — 1) 

r =r -. 

e cos. q> -\- 1 

If we take 

P — FC = c — A = A (e — 1), 



(81) 



<§> 98.] EQUATIONS OF LOCI. 67 

Rectangular equation of hyperbola. 



these may be put in the form 

P(l+e) 

r := = 

e cos. cp — 1 



(82) 



r=- P(l + ' ) .. (83) 

e cos. (p -\- 1 ' 

Corollary 9. If we draw ECE perpendicular to CC' t 

and make *-4E* ~»^ 

CE = C£ ; = ^F = A F< — A 

we have 

AC c 
cos. jELIC = cos. E'AC = — = = —, 

— COS. (p Q . 

Hence -EMC = E'AC = <P 0> 

and JJj^C^ztt: — y Qi 

E X AC =Z TV -f- (po- 

CE — */(AE 2 — AC 3 ) = \/(A 2 — c 2 ) = B. 

Corollary 10. The equation of the hyperbola may be re- 
ferred to rectangular coordinates by arts. 59 and 60. But 
since equation (76) differs from the equation (64) of the 
ellipse only in regard to the value of c, this equation may be 
referred to the rectangular axes CAC and BAB 1 , by the very 
same formulas as in corollary 6 upon the ellipse, and we 
shall have 

(A 2 — c 2 ) x 2 +A 2 y 2 = A 2 (A 2 — c 2 ), 
or, substituting B 2 , 

— B 2 z 2 +A 2 y 2 =z — A 2 B 2 , (84) 

which, divided by — A 2 B 2 , is 

%2 y 2 - 1 tan 

A?~B*- L (85) 






68 ANALYTICAL GEOMETRY. [b.I. CH.IT, 

Hyperbola referred to oblique axes. 

With regard to equation (77), since it may be deduced from 
equation (76), by changing c into — c, or — c into c, it may 
be referred to rectangular coordinates by the same process, 
and the corresponding result may be deduced by changing in 
that for (76) c into — c. Since, however, 

C 2 = (—c) 2 , 

the result is the same in both cases. 

Equations (84) and (85) are, then, the equations of 
both branches of an hyperbola referred to rectangular 
coordinates, the centre of the hyperbola being the origin, 
the transverse axis being the axis of x, and the conju- 
gate axis being the axis of y. 

Corollary 11. If we wished to find the point where the 
curve meets the axis of y, we should have for these points 

x — 0, 

so that the corresponding value of y would be 

(^4 2 i3 2 \ 
jr 2 -) = */-B* = ±Bs/-\, 

which is imaginary, and there are no such points. 

Corollary 12. The equation of the hyperbola may, by art. 
71, be referred to oblique axes. If the origin remains at A, 
the result is the same as that of corollary 9 for the ellipse, by 
changing B 2 into — B 2 . By this change (70) becomes 

(A 2 sin 2 a — B~ cos. 2 «)x 2 -|- 2(^1 2 s i n . a sin./S — JB 2 cos. « cos.,?)* x y x 

+ (A 2 sin. 2 /* — B 2 cos- 2 (*) y\ = — A 2 B 2 . (86) 



§ 98.] EQUATIONS OF LOCI. 69 

Hyperbola referred to conjugate diameters. 

Corollary 13. If <* and p are so taken, that 

A 2 sin. a sin. P — I? 2 cos. « cos. /? = 0, (87) 

I? 2 

or tang. « tang. ? =- — (88) 

the axes are said to be conjugate to each other ; and equation 
(86) becomes 
(4 2 sin. 2 a- J B 2 cos. 2 «)x 2 +(^L 2 sin. 2 / 5-J5 2 cos. 2 1 5)?/ 2 =— ^ 2 jB 2 (89) 

Corollary 14. It may be proved precisely as in corollary 11 
for ellipse, that a line drawn through the centre to meet the 
curve at both extremities is bisected at the centre, whence it 
is called a diameter. If such a diameter is assumed for the 
axis of x, and if we denote it by A 1 , we have 

A ,= AB 

\/(B 2 cos. 2 « — JL 2 sin. 2 «) 

and if we take 

AB 



B 



\/(A 2 sin. 2 ? — B 2 cos. 2 /*) 
we have 

A 2 B 2 

A 2 sin. 2 a — B 2 cos. 2 p == , 

A 12 ' 

A 2 Tt 2 

A 2 sin. 2 p — B 2 cos. 2 /5 — - -^ - , 

which substituted in (89) give, by dividing by — A 2 B 2 , 

T 2 II 2 

or 

— B l2 %\ + A l2 y 2 — —A 2 B' 2 i (91) 

which are the equations of the hyperbola referred to 
conjugate diameters. 



70 ANALYTICAL GEOMETRY. [b. I. CH. IV. 

Parabola. Polar equation. 

5. Find the equation of the locus of all the points 
in a plane so situated, that the distance of each of them 
from a given point is equal to its distance from a given 
line. This locus is called the parabola, the given point 
its focus, and the given line its directrix. 

Solution. Let the given point be (fig. 37.) assumed as the 
origin of polar coordinates, and let the perpendicular AF to 
the given line EQ be produced to X, and let FX be the 
polar axis. Let 

B F = 2 P. 
Draw the perpendicular MP ; we have 

r — FM— Q31 = BP 

FP z= r cos. q>, 
so that 

r cos. <p = BP — BF —r — 2P 

r — r cos. (p — 2 P 

2 P 
r = -r-^- , (92) 

1 C os. cp 

which is the polar equation of the parabola, the origin 
being the focus, and the axis the perpendicular from 
the directrix. 

Corollary 1. If equations (67) and (92) are compared to- 
gether, it is evident that (92) is what (67) becomes, when 

e— 1. 

Corollary 2. For the point A where the curve meets the axis 
we have 

(p — TV , COS. (p = — 1 

r — FA — \1P — P 



<§> 98.] EQUATIONS OF LOCI. 71 

Parabola referred to rectangular axes. 

The point A is called the vertex of the parabola, and is 
just as far from the focus as from the directrix. 

Corollary 3. The equation of the parabola may be referred 
to rectangular coordinates by arts. 59 and 60. If we take 
the vertex A for the origin, we have 

a ~ FA — P 

a = 0, $ •= n, 

sin. (? — «) == 0, cos. (/* — o) == — 1 ; 
whence (9) and (10) become 

r = s/(x 2 -\-y 2 — 2Pz + P 2 ) 

tang. 9 = ^p 
1 x — P % — P 



COS. 



a/ (I +tan. 2 c P )~ /V /( j/ 2_|_ a; 2_ 2Px+P 2 ) 



r cos. 



— x — P. 



Now equation (92), freed from fractions, is 

r — r cos. ? = 2P, 
in which, if we substitute the preceding values, we have 
^/( y 2 _|_ a;2 _ 2 P a: + P 2 ) — x + P = 2 P 
\/(*/ 2 + z 2 — 2Px-\- P 2 ) = P + x; 
which squared and reduced gives 

f = 4Pa;; (93) 

which is the equation of the parabola referred to rect- 
angular coordinates, the origin being the vertex, and P 
its distance from the focus. 



72 ANALYTICAL GEOMETRY. [b. I. CH. IV. 

Parabola referred to oblique axes. 

Corollary 4. The equation of the parabola may be referred 
to oblique axes, by art. 71. If the axis of x x is taken parallel 
to x, we have 

« = 0, sin. a = 0, COS. « = 1, 
and (27) and (28) become 

x = a-\-x x +y 1 cos. /s 

V — hJ rVi sin - P'> 
which, substituted in (93), give 

y 2 sin. 2 /? + (2 b sin. p — 4Pcos.P)y 1 

+ 62 -4Pa= 4Pz 1 . (94) 

Corollary 5. If the new origin is taken at a point A x upon 
the curve, we have, by equation (93), 

6 2 = 4Pfl, 
which reduces (94) to 

y\ sin. 2 i* -L. (26 sin. /s— 4 P cos. (?)y 1 = 4Pz 1 (95) 

Corollary 6. If the inclination (9 is taken so that 
2 6 sin. p — • 4 P cos. ?z:0, 

2P 

or tan. p =z — -, (96) 



(95) becomes 



And if we put 



(98) becomes 



y\ sin. 2 |5z= 4Pz x (97) 

4P 

sin. 2 P * ' 



rt^J£?*i> ( 98 ) 



p 

Pl = ■ o > (") 

1 sin. 2 /s ' - v J 

y? = 4P 1 x 1 . (100) 



§ 98.] EQUATIONS OF LOCI. 73 

Prolate ellipsoid of revolution. 

The axes, determined by the equation (96), are said 
to be conjugate to each other, and (100) is the equation 
of the parabola referred to conjugate axes. 

6. To find the equation of the surface described by 
the revolution of the ellipse about its transverse axis. 
This surface is called that of the prolate ellipsoid of 
revolution, which is the included solid. 

Solution. Let CMC (fig. 34.) be the revolving ellipse. 
If we use the notation of the 3d problem and solution, and 
let F be the origin of the polar coordinates in the system of 
art. 73, and the axis of revolution the polar axis, it is evident 
that the distance F31 of each point from the origin, or any 
other point of the axis, remains unchanged during the revolu- 
tion of the ellipse. The value of r is then independent of 0, 
and depends only upon the angle y, which it makes with the 
axis. Hence the equation (64) of the ellipse determines the 
value of r for every value of <p and every position of the re- 
volving ellipse. 

It is, then, the polar equation of the prolate ellip- 
soid. 

Corollary 1. The equation of the ellipsoid may, by art. 77, 
be referred to rectangular coordinates. Thus, if in equation 
(64) freed from fractions we substitute 

r = V(z 2 +3/ 2 + * 2 ) 
r cos. <p z=z z, 
we have 

Aa % /(x2 -f-y2 _j_ 2 2) _ CZ= i2- C 2 (101) 

r 



74 ANALYTICAL GEOMETRY. [b. 1. CH. IV. 

Ellipsoid of revolution. 

Corollary 2. This equation may, by art. 90, or 93, be re- 
ferred to other rectangular axes. Thus, if the origin is 
changed from JF to A, we have for the a, b, c, of art. 93, 

a = FA = c, 6 = 0, c = 0, 
whence 

x — x i + c > y — y^ 2==*ii 

which, substituted in (101), give 

^/[(*i+«) 2 +y?+*?]=^ 2 — c 2 +«(*i+0=^ 2 +'*i- 

Squaring and reducing, we have 

(A 2 — c 2 )x\-\-A 2 (y 2 + z 2 ) — A2 (A 2 — c 2 ) ; 
and substituting the B 2 of corollary 4 of the ellipse 

B 2 x\ + A 2 y 2 + A 2 z 2 — A 2 B 2 , (102) 

which, divided by A 2 B 2 , is 

nr-2 1/2 2 2 

f*+W + m = '' < ,03 > 

which is the equation of the prolate ellipsoid of revolu- 
tion referred to its centre as the origin, the axis 2 A of 
revolution being the axis of x x . 

7. To find the equation of the surface, generated by 
the revolution of the ellipse, about its conjugate axis. 
This surface is that of the oblate ellipsoid of revolution. 

Solution. If we take the centre A (fig. 38.) of the ellipse 
whose transverse axis is 

CC = 2 A, 
and conjugate axis 

B B' = 2 B 



§ 98.] EQUATIONS OF LOCI. 75 



Oblate ellipsoid of" revolution. 



for the origin of rectangular coordinates ; the equation of this 
ellipse is 

ll _L Hi — 1 

A? "t" B* ~~ 

When it revolves about the axis BB' the distances MR and 
AR remain unchanged. Let x lt y lf z x be the coordinates of 
the point 31 of the required surface, BAB' being the axis* of 
sEj. We have 

AR = x x . 

Now the distance of the point M from the point P is, by 

art. 78, 

MR^^[(x 1 -x 1 )2 + y2 +z 2 ] = x/{y 2 +z 2y 

But MR and AR are the same with coordinates AP and 
MP, or x and y of the point M in the plane of the ellipse ; 
so that, for this point, 

y = x xi x = MR=*/(y2+z2), 

which, substituted in the equation of the ellipse, give 

1+3 + 1=1; d04) 

which is the equation of the oblate ellipsoid of revolution 
referred to its centre as the origin, the axis 2 B of revo- 
lution being the axis of x v 

8. To find the equation of the surface formed by the 
revolution of the hyperbola about either its transverse 
or its conjugate axis. This surface is that of the 
hyperboloid of revolution. 



76 ANALYTICAL GEOMETRY. [b. 1. CH.IV. 

Hyperboloid and paraboloid of revolution. 

Solution. By reasoning exactly as in the preceding solution, 
we find 

f!_ll_*- = i (105) 

42 B 2 B 2 K ' 

for the equation of the hyperboloid of revolution referred 
to its centre as the origin, the transverse axis 2 A being 
the axis of x and also the axis of revolution, 

and we find 

x 2 v 2 z 2 

-z^+i* + -^ = l < 106 > 

for the equation of the hyperboloid of revolution referred 
to its centre as origin, the conjugate axis 2 B being the 
axis of x, and also the axis of revolution. 

9. To find the equation of the surface generated by 
the revolution of the parabola about its axis. This 
surface is that of the paraboloid of revolution. 

Solution. By reasoning exactly as in the preceding solu- 
tions, we find 

y 2 -\-z 2 = 4Px (107) 

for the equation of the paraboloid of revolution referred 
to its vertex as origin, the axis of revolution being the 
axis of x. 

10. To find the equation of the straight line in a 
plane. 

Solution. Let AB (fig. 39.) be the line, let any point A in 



§ 98.] EQUATIONS OF LOCI. 77 

Straight line. 

it be assumed as the origin of polar coordinates, and let the 
polar axis be AX, which is inclined to BA by the angle 

BAX — x. 

For every point M of this line we have, then, 
9 = MAX— 2; 

so that tf == x (108) 

is the polar equation of a straight line, which passes 
through the origin, and is inclined to the axis by the 
angle *. 

Corollary 1. The equation of the axis is 

<P = 0. 

Corollary 2. The straight line may be referred to rectan- 
gular axes by art. 60, and if the axis of x is that of AX, 
(11) gives 

tang, x =z ■¥-, (109) 

or y = % tang, x ; (1 10) 

which is the equation of the straight line, which passes 
through the origin, and is inclined to the axis of x by 
the angle *. 

Corollary 3. For the axis of x 
x = 0; 
so that y = (111) 

is the equation of the axis of x. 
7* 



78 ANALYTICAL GEOMETRY. [b. I. CH. FT. 

Straight line. 

In like manner 

# = (112) 

is the equation of the axis of y. 

Corollary 4. The straight line may be referred to any 
oblique coordinates by art. 71. But since the new axes may 
be situated in any way whatever with regard to the former 
ones, the generality of the result is not diminished by limiting 
the original position of the line to that of the axis of y, cor- 
responding to equation (112). 

Thus, if the new origin is at the point A x , we have 

a = A'A — A"A iy 
and (27) becomes 

x x cos. a -\- y 1 cos. /? = — a. 
Now — a is the value of A"A 1 counted from A lt or it is 
the perpendicular let fall upon the line from the new origin, 
and if we put 

p — —a, 

we have 

x x cos. a -\- y x cos. /} = p ; (1 13) 

which is the equation of a straight line passing at the 
distance p from the origin, « and & being the angles 
which the perpendicular to the line makes with the axes 
of x x and y \- 

Corollary 5. Equation (113) may be applied to the case in 
which the new axes are rectangular, when 



3 



cos. P = — sin. « and cos. «- — sin. p, 



§ 98.] EQUATIONS OF LOCI. 79 

Straight line. 



and (113) becomes 

cos. a.x 1 — sin. «.y 1 =p, (114) 

or — sin. |S. x x -\- cos. /J. y x = p (115) 

cos. j$ . y x =3 Xj sin. i 5 +p 

y 2 == x x tan. I* +p • sec. p, 

in which /S is the angle made by the line itself with the axis 
of ** 

In fig. 40 let AB be the line, we have in the right triangle 
A X PB, formed by letting fall the perpendicular A X P, 

A x P=p y PA X B^=? 

A X B = A x P sec. PA x B =s p sec. /?, 
and, if 

A == JL x B — p sec. (S 

y x —x x tang. /» + h, (116) 

which is the equation of a straight line inclined to the 
axis of x x by the angle ?, and cutting the axis of y x at 
a height h above the origin. 

Corollary 6. The equation of the straight line may be ob- 
tained for any polar coordidates by art. 47, or more simply by 
art. 61, applied to the axis of y ; in this case we have, as in 
corollary 4, 

a — —p, 
and (12) substituted in (112), gives 

r cos. (g> + u ) = P> (11?) 

which is the polar equation of a straight line passing 
at the distance p from the origin, the perpendicular 
upon it being inclined to the axis by the angle — «. 



80 ANALYTICAL GEOMETRY. [b. I. CH. IV. 

Straight line in space. 

11. To find the equation of a straight line in space. 

Solution. If a point in the line is assumed as the origin, and 
such rectangular axes of x, y, z, that the straight line makes 
with them the angles A, u y Vf the polar equations of the line 
in the system of art. 87, are 

<p = *, y = ft, w — v . (118) 

Corollary 1. It must not be forgotten that A, fli v are not 
entirely independent of each other, but are subject to the re- 
striction of art. 86, 

cos. 2 x -f- cos. 2 /t + cos. 2 v = 1. (H9) 

Corollary 2. The equations of the axis of x are 

9> = 0, v = J w , w = l'»5 (120) 

those of the axis of y are 

. y = Jtt, y = 0, w = J*r; (121) 

those of the axis of z are 

CP = J 7T, y = j 7T, W — 0. (122) 

Corollary 3. Equations (39) become, by substituting in 
them the values of y, V, w (118), 

X x — r cos/\ A 

U tt 

y — r cos. v 

I z — r cos. X ; 

whence 

r= — =-^ = — , (123) 

cos./* cos.* cos. a : 

which are the equations of a straight line passing 



§ 98.] EQUATIONS OF LOCI. 81 

Plane. 

through the origin, referred to rectangular coordi- 
nates. 

Corollary 4. The equation of the straight line may be re- 
ferred to any rectangular axes by equations of art. 93, by 
which (123) becomes 

',+« = y 1 +i s *, + c (124) 

COS. a COS. " COS. v 

which are the equations of a straight line which passes 
through the point, whose coordinates are — a, — -b, — c, 
and is inclined to the axes by the angles ;., ,", »-. 

12. To find the equation of the plane. 

Solution. If the plane is that of xy. we have for all 
its points, by art. 74, 

z = 0, (125) 

which is, then, the equation of the plane xy. 
In the same way 

y = o . (126) 

is the equation of the plane x z : also 

x = (127) 

is the equation of the plane y z. 

Corollary 1. The plane may, by arts. 90, 91, be referred to 
any axes whatever. Thus, for the plane of y z, equation (40) 
gives, by (127), 

x x cos. « + y 1 COS.0 + Z, cos. r = — «*== jp (128) 

which is therefore the equation of the plane, which 



82 ANALYTICAL GEOMETRY. [b. I. CH. IV. 

Cycloid. 

passes at the distance — a or p from the origin, and 
the perpendicular to which is inclined by the angles 
«, i*, r, to the axes x lt y lt z 4 . 

Corollary 2. If the plane passes through the origin, we 
have 

P = 0, 

and (128) becomes 

x x cos. « -|- y x cos. p -|- z x cos. yz:0, (129) 

13. To find the equation of the curve described by 
a point in the circumference of a circular wheel, which 
rolls in a plane upon a given straight line. This curve 
is called the cycloid. 

Solution. Let the given straight line AX (fig. 41.) be the 
axis of x, and let the point A t at which the given point M of 
the circumference touched AX, be the origin ; let 

R = the radius CM of the wheel, 

6 = the angle MCB, by which the point M 
is removed from B. 

Then, since the arc BM has rolled over the straight line AB J 
it must be equal to it in length, or, by art. 97, 

R6 = BIB — AB. 

Draw ME parallel to AX; the right triangle CME gives 

CE — R cos. 6, ME = BP — R sin. 6, 
whence 

x = AP = AB — BP— R& — 22siD.fi, (130) 

y= PM— BE—CB — CE = R — Rcos.6, (131) 



<§> 98.] EQUATIONS OF LOCI. 83 

Cycloid. Spiral. 

and the elimination of 6 from these two equations would give 
the required equation. This elimination is thus effected ; (131) 
gives, by transposition, 

R cos. & = R — y, 
whence 

R sin. 6 — s/{R 2 — R2 cos. 2 6) — ^[R2 _ (72 — y)2] 
which, substituted in (130), gives 

X = R& — s/iflRy— y2) 

_ x + V(2 Ry— y2) 
& R~ • 

which, substituted in (131), gives 

, = R-B«*( * + ''P]l'->') ) (132). 

which is the equation of the cycloid, but is not so coti- 
venient for use as the combination of the two equations 
(130) and (131). 

14. A line revolves in a plane about a fixed point of 
that line, to find the equation of the curve described by 
a moving point in that line, which proceeds from the 
fixed point at such a rate, that its distance from the 
fixed point is proportionate to the 7ith power of the 
angle made by the revolving line with the fixed line 
from which it starts. This curve is called a spiral. 

Solution. Let the fixed point A (fig. 42.) be the origin, and 
the fixed line AB the polar axis. Let M be the moving point 



84 ANALYTICAL GEOMETRY. [b. I. CH. IV. 

Spirals. 

which, after the line has revolved completely round once, has 
arrived at M'. Let 

R = AM', 
we have, by condition, 

r : R z=z <p n : (2 ?*)», 
or r ( 2 tt)» = 22 9> B , (133) 

for the equation of the spiral. 

Corollary 1. If n = 1 and 22 = 1, 

equation (133) becomes 

2 7rr=9, (134) 

which is the equation of the spiral of Conon or of Ar- 
chimedes. 

Corollary 2: If n — — 1, 

(133) becomes 

(2 7r)-i r = jR<p~i, 
or rcp=z2nR } (135) 

which is the equation of the hyperbolic spiral. 

Corollary 3. If the logarithm of the distance of the 
point had been equal to the angle, the curve would 
have been the logarithmic spiral, and its equation 

<f s= log. r. (136) 

15. To find the equation of the right cylinder, whose 
base is a circle. 

Solution. Let the plane of the base be that ofxy. For 
any point whatever (fig. 43.) the coordinates of its pro- 



§ 98.] EQUATIONS OF LOCI. 85 

Right cylinder. Right cone. 

jection P upon the plane of x y are x and y. But since the 
point P is in the circumference of the base, x and y must 
satisfy the equation of this circumference. 

The equation of the right cylinder is then the same 
as that of its base, if the plane of this base is assumed 
as one of the coordinate planes. 

Corollary. The preceding proposition is obviously 
general, and may be applied to any right cylinder what- 
ever, be its base a circle, an ellipse, an hyperbola, or 
any other curve. 



16. To find the equation of the right cone, whose 
base is a circle. 

Solution. Let the vertex A (fig. 44.) be assumed as the 
origin, and let the axis of x be that of the cone, and let 

X — the angle which the side of the cone makes 
with the axis. 

Every radius vector, as AM, is a part of a side, and there- 
fore 

cp =z x (137) 

is a property of every radius vector, and is the polar 
equation of a right cone, the origin being the vertex, 
and i being the angle made by the side with the axis. 

Corollary 1. The equation of the cone may, by art 88, be 
referred to rectangular axes, and we have 

x x 
COS. w z=. — zzr — ; zz: cos. *. 

9 r \/(x 2 +y 2 +% 2 ) ' 

8 



86 ANALYTICAL GEOMETRY. [b. 1. CH.IV. 

Right cone. 

so that 

x = s/{%* + y* _|_ %2 ) C0S- a (138) 

is the required equation. 

Squaring and transposing, we have 

(y 2 + z2 ) C0S - 2 2 = x2 (! — " C0S - 2 x ) — * 2 sin. 2 a 

3/2 _|_ %2 = Z 2 tang. 2 a, (139) 

which is the equation of a right cone whose vertex is 
the origin, and axis the axis of x. 

Corollary 2. The equation of the cone may be referred 
to any rectangular axes which have the same origin, and 
which make the angles a, /s, y with the axis of x, by arts. 90 - 
92 ; for we have 

r=:V(x2+y2 +z 2) ==(V /( x 2 + ^2 +%2) 

X = X 1 COS. a + V x COS. /S -\- % x COS. y J 

which, substituted in (138), give 

x 1 cos.a + yiCOs.|»+z 1 cos. y =r,v/( a; ?+y?+z?)cosA (140) 

Corollary 3. If the origin is changed to the point, whose 
coordinates are a, b = 0, c = 0, the equation becomes 

{x 2 -\- a) cos. a ~\-y 2 cos « Z 5 + z 2 cos - y 

= v'[(^ + «) 2 +yl + ^]co3.i. (i4i) 



§ 102.] ORDERS OF CURVES. 87 

Order of curves. 



CHAPTER V. 

CLASSIFICATION AND CONSTRUCTION OF LOCI. 

99. When the equations of loci are referred to rect- 
angular coordinates, they are divided into degrees, or 
orders, corresponding to the degree of their equation. 
Thus the locus, whose equation is of the first degree, 
is itself of the first degree or linear, and the same is 
the case with other curves. 

100. Theorem. The order of a curve is independent 
of the particular system of rectangular coordinates to 
which it may be referred; that is, it is of the same 
order for all systems of rectangular coordinates. 

Demonstration. The formulas (16, 17) or (40, 41, 42) 
for transforming from one system of rectangular coordinates 
to another are linear ; so that the greatest number of the di- 
mensions of .t, y, z in any term must be the same with that of 
the dimensions of x t , y lf z v The degree of the equation is. 
therefore, the same, when expressed in terms of x xi y 1} z lf 
that it is when expressed in terms of x, y, z. 

101. Corollary. Since the equations for transforming 
to oblique coordinates are also linear, the preceding 
proposition may be extended to them. 

102. Corollary. The degree of the circle is, by (58), 



88 ANALYTICAL GEOMETRY. [b. I. CH. V. 

Construction of loci. 

the second ; likewise that of the sphere (62) ; of the 
ellipse (69) ; of the hyperbola (85) ; of the parabola 
(93) ; of the cylinder and of the cone. The degree of 
the straight line (123) is the first, or it is linear, as is 
also that of the plane (128). The equations of the 
cycloid, and of the spirals, cannot be expressed without 
the aid of arcs, so that these curves are transcenden- 
tal. 

103. Problem. To construct a locus, of which the 
equation is given. 

Solution. I, If the equation is that of a locus in a 
plane, and expressed by polar coordinates, we can, by 
giving successive values to q>, differing but little from 
each other, calculate, by means of the given equation, 
the corresponding values of r. As many points of the 
required curve may thus be determined as may be 
convenient, and the curve, which is drawn by the 
hand through these points, cannot differ much from 
the required curve. 

II. If the locus were in a plane and expressed by 
rectangular coordinates, points might be determined by 
calculating for assumed values of x the corresponding 
values of y. 

III. If the equation were that of a locus in space, 
and expressed by polar coordinates ; then for each as- 
sumed direction of the radius its value might be calcu- 
lated, and the locus obtained by joining the series of 
points thus determined would obviously be a surface. 



<§> 105.] ORDERS OF CURVES. 89 

Construction of loci. 

IV. If the equation were that of a locus in space, 
expressed by rectangular coordinates, values might be 
assumed for x and y, and the corresponding value of z 
would express the height at which the point of the 
locus was above its projection upon the plane x y ; so 
that this locus would also be that of a surface. 



V. If there were two equations in space, then one 
of the coordinates might be assumed at pleasure, and 
the corresponding values of the other two obtained. 

104. Corollary. A single equation between coordi- 
nates in space denotes a surface. But if there are two 
equations, the coordinates of each point of the locus 
must satisfy each equation, and the point must be at 
once upon both the surfaces represented by these equa- 
tions ; so that the locus is the intersection of these 
surfaces, and is consequently a line. 

105. In determining .the character of loci from their 
equations, it is important that these equations should be 
first of all referred to those coordinates, for which they 
are the most simple in their forms. 



8* 



90 ANALYTICAL GEOMETRY. [b. I. CH. VI. 

Reduction of linear equation. 



CHAPTER VI. 

EQUATION OF THE FIRST DEGREE. 

106. The general form of the equation of the first 
degree in a plane is 

Ax + By + M =0, (14£) 

and that of the first degree in space is 

Ax + By + Cz + M= 0. (143) 

107. Problem. To reduce the general equation of the 
first degree in a plane to its most simple form. 

Solution. Let the general formulas (16) and (17) for trans- 
formation from one system of rectangular coordinates to an- 
other in a plane be substituted in the general equation (142). 
The result is 

(A cos. a-\-B sin. «) x t -\- ( B cos. « — A sin. «) y x 

+ Aa+Bb + M = (144) 

in which a and b are the coordinates of the new origin, and 
« the angle made by the axes x and x x . 

Now the position of the new origin may be assumed at such 
a point that 

Aa + Bb-\-M-z 0, (145) 

and the angle « may be so assumed that 

A cos. a -J- B sin. « = 0, 

or tang. « = =, (*46) 



§ 109.] LINEAR LOCUS. 91 

Linear locus. Perpendicular. 

whence 

1 B 

cos. « — 



V(l + tang.* a) _ V^ 2 + B 2 ) 

A 

SID. a = COS. cc . tang. « zrr — 



V(^ 2 + B 2 )' 
and (144) is reduced to 

A 2 +B 2 _ 

s/(A 2 + B*) Vl T 

or V(^ 2 +^ 2 )yi = o, 

whence y x = 0, (147) 

which is as simple a form as the given equation can attain. 

108. Corollary. Since 

is, by (111); the equation of the axis of x 1} £Ae locus 
of the given equation is a straight line, which passes 
through the point of which the coordinates are a and b, 
and is inclined to the axis of x by the angle whose 
tangent is — A -j- B. 

109. Corollary. If the given equation (142) is divided by 
\/(A 2 -\- B 2 ), and cos. « and sin. « are substituted for their 
values, it becomes, by transposition, 

-sm. g .x + cos.«.y = - V ( J% 3) , 
which, compared with (115), leads to the result that 

M 
\/(A 2 +B 2 ) 

is the length of the perpendicular let fall upon the line 
from the origin. 



92 ANALYTICAL GEOMETRY. [b. I. CH. VI. 

Angle of two lines. 

Of the two values of 

V(-4 2 + £ 2 ) == ±V(^ 2 + £ 2 ), 
that value should then be taken which renders 

_ M ~- V(^ 2 + B2 ) 

positive 3 that is, the value which is of the same sign 
with — M. 

110. Problem. To find the angle of two lines in a 
plane, whose equations are given. 

Solution. Let their equations be 

and let a, a x be the angles which they respectively make with 
the axis of x ; we have for the angle I, which they make with 
each other, 

I=a x —a. (148) 

But, by (146), we have 



and 



A A 

tang, a — — — , tang. « x = — -j± 



tang. I = tang. (a 1 — «) 



tang. « j — tang, a 
~ 1 -|- tang. «tang. « 2 

tang . 7= i|^_^. {149 ) 



§ 114.] LINEAR LOCUS. 93 

Parallel and perpendicular lines. 

111. Corollary. If the two lines are parallel, we have 

1=0 
tang. 7=0 
A 1 B — B 1 A = 0, (150) 

or %=i (151) 

for the equation expressing that the two lines are paral- 
lel. 

112. Corollary. If the two lines are perpendicular, we 
have 

/= Jtt, 

tang. I = oo = $ ; 
or the denominator of (149) must be zero; that is, 

AA 1 + BB 1 ~0 (152) 

is the equation expressing that the two lines are perpen- 
dicular. 

113. Corollary. In case the two lines are parallel, their 
distance apart must, by art. 109, be 

M M x 



s/(A2+B*) ' V(A1+B*)' 

114. Problem. To find the coordinates of the point 
of the intersection of two straight lines in a plane. 

Solution. Let the coordinates of the point of intersection 
be x and y Qf and let the equations of the line be the same 
as in the preceding article. Since the point of intersection is 



94 ANALYTICAL GEOMETRY. [b. I. CH. VI. 

Intersection of two lines. 

upon each line, its coordinates must satisfy each of their equa- 
tions, or we must have 

A x + B y + M == 0, 
A^ + B.y.+M^O; 
from which the values of x and y are found to be 
BM.-B.M 
*°- AB.-A.B' (153) 

A,M—AM 1 .. 

y^ AB.-A.B ' (154) 

115. Corollary. If the equations of the line had been given 
in the form corresponding to (115) 

— sin. « . x -j- cos. a y z=i p 

— sin. a 1 .z + cos.« 1 y — p x 

we should have found 

p COS. « x p i COS. a _pCOS.« 1 ^COS^ 

sin. « x cos. « — .cos. a x sin. a sin. (« x — «) 

^sin. a,— p .sm.a 
y °— sin. («! — «) ' 

116. Corollary. The values of z and y (153) and (154) 

would be infinite, if their denominators were zero, that is, if 

we had 

AB X — A^zzzzO, 

or by (150) if they were parallel, in which case they would 
not meet, and there would be no point of intersection. 

117. Problem. To find the equation of a straight 
line, which makes a given angle with a given straight 
line. 



§ 120.] LINEAR LOCUS. 95 

Line inclined to given line. 

Solution. Let the given angle be 1, and the equation of the 
given straight line 

— sin. a . x -\- cos. a . y — : p } 
and let that of the required straight line be 

— sin. cfj . x -\- cos. a 1 ,y = p 1 , 

in which <* 1 and p x are unknown. We have, by the condi- 
tions of the problem, 

«! — « — J, orsj = I-\- «; 

and this value of ffl , being substituted in the equation of the 
required line, gives 

— sin. (I-\- a ). x -f- cos. (J-f- «) y = p 1 (157) 

for the required equation, in which p x is indeterminate, be- 
cause there is an infinite number of lines which satisfy the 
condition of the problem. 

118. Corollary. If the required line is to be parallel to the 
given line, we have 

1=0, 
and (157) becomes 

— sin. a.x -j- cos. a.y = p x . (158) 

119. Corollary. If the required line is to be perpendicular 
to the given line, we have 

/ = i™, sin. ( I -f- «) = cos. «, cos. (7+ «) = — sin. «, 

and (157) becomes 

cos. a x -\- sin, ay — — p x . (159) 

120. Problem. To find the equation of a straight 
line, which passes through a given point. 



96 ANALYTICAL GEOMETRY. [b. I. CH. VI. 

Line passing through given points. 

Solution. Let x', y' by the coordinates of the given point, 
and 

— sin. a . x -\- cos. « . y — p, 

the required equation in which a and p are unknown. Since 
the given point is in the required line, its coordinates must 
satisfy this equation, and we have 

— sin. a . «' + cos * « • V' — P> (!60) 

which is a condition that must be satisfied by « and p ; al- 
though it is not sufficient to determine their values, because 
many different lines can be drawn through the same point. 
If the value of p is substituted in the required equation, it 
becomes, by transposition, 

— sin. a . (x — x') -\- cos. «. (y — y') — 0, (161) 

or, dividing by cos. a, 

— tang, a (x—x') + (y — y>) = 0, (162) 

which is the required equation, « being indeterminate. 

121. Corollary. If this straight line is also to pass through 
another point, the coordinates of which are x" and y" } we also 
have this condition corresponding to (160) 

sin. a , x" -f- COS. « . y" — p f 

from which and (160) the values of p and « are to be found. 

The difference between the last equation and (160), divided 
by cos. a and by x" — x' ', is 

tang- • = f^| ; (163) 

which substituted in (162) gives, by transposition, 

y-y>=J!^^(z-Z<) (164) 



<§> 124.] LINEAR LOCUS. 97 

Parallel and perpendicular to given line. 

for the equation of a straight line, which passes through 
the two points whose coordinates are x' , y' and x" , y". 

122. Corollary. If the straight line of art. 120 has also to 
make a given angle with the straight line whose equation is 

— sin. «i x -\- cos. a 1 y — p 

we have, by art. 117, 

which substituted in (162) gives, by transposition, 

y-y' = tang. (/+ «J (x — x>) (165) 

for the required equation. 

123. Corollary. If the two lines of the preceding article 
are to be parallel, we have 

1=0, 

and (165) becomes 

y —y> = tang. a x (x — x'). (166) 

124. Corollary. If they are to be perpendicular, we have 
■T=i^; tang. (J^ + a x ) — — cotan. « 1} 

and (165) becomes 

y —y 1 — — cotan. « 1 ( x — z'), (167) 

which is, therefore, ?Ae equation of the perpendicular 
let fall from the point, whose coordinates are x' y' upon 
the straight line, whose equation is 

— sin. «j x -J- cos. «j y — 2^. 

9 



98 ANALYTICAL GEOMETRY. [b. I. CH. VT. 

Length of perpendicular to line. 

125. Corollary. For the point of intersection of the per- 
pendicular (167) with the line upon which it falls, that is, for 
the foot of the perpendicular, we find by the process of art. 
114, 

x Q — y' sin. ai cos. a 1 -f- z'cos. 2 a x — p x sin. Ul (168) 
y = y'sin. 2 « x -[-a;' sin. a 1 cos.a 1 -\-p ± cos. a yl ; (169) 

126. Corollary. The length p of this perpendicular is 
the distance between the points x' } y and x ,y , so that by 
equation (23) 

p% = (*' — *o) 2 + (y' — yo) 2 > 

but by (168) and (169) 

x' — x — x'(l — cos. 2 «j) — y's'm.a 1 cos. « x -{-p 1 s'm. a 1 
= x' sin. 2 a x — y' sin. a 1 cos. a x -\-p\ sin. « x 
= {x 1 sin. «j — y' cos. « 2 -}- p t ) sin. ai 
y' — y == — (x' sin. « x — y cos. % +^ x ) cos. « lf 
so that 
p\ = (x'sin. «! — y' cos. a x -{-p^ 2 (sin. 2 « x -|- cos. 2 « x ) 

= (x'sin. a x — /cos. «j +Pj) 2 
|) r= x' sin. a x — y' cos. « 2 +p x . (1^0) 

127. Problem. To reduce the general equation of 
the first degree in space to its most simple form. 

Solution. Let the general formulas (40, 41, 42) for trans- 
formation from one equation of rectangular coordinates to 
another in space be substituted in the general equation (143) 

Ax+By + Cz + M= 0, 



<§> 128.] LINEAR LOCUS. 99 

Linear locus in space. 
the result is 

(A COS. « -f- B COS. a' -)- C COS. a") x x 

■ + (A cos. ? + B cos. (?' + C cos. |S") y 
-f- (J. cos. 7 + -B cos. / -|- C cos. 7") z x 

+ ifl + 56+Cc + i/zzO, (171) 

in which «, #, 7, «', ^ y 1 , a", p", y" are subject to the six con- 
ditions (44-49). 

Let now the position of the new origin be assumed at such 
a point, that its coordinates a, b, c satisfy the equation 

Aa-}-Bb + Cc + 31 — 0, (172) 

and let the angles ? and y be subject to the two conditions 

A cos. ? + B cos. ?' +C. cos. <*" =: (173) 

A cos. 7 + B cos. / + C. cos. y" = 0. (174) 

By this means equation (171), divided by 

A COS. a -\- B COS. a' -J- C COS. a", 

is reduced to 

x x = 0. (175) 

128. Corollary. Let 

^ cos. a-\-B cos. «' + C cos. «" — Z, (176) 

and if (176) is multiplied by cos. «, (173) by cos. p, (174) by 
cos. y, the sum of the products, reduced by means of equa- 
tions (47,50, 51), is 

A = L cos. «. (177) 



100 ANALYTICAL GEOMETRY. [b. I. CH. VI- 

Equation of plane. 

In the same way we find 

B = L cos. «' (178) 

(7=icos. «". (179) 

The sura of the squares of (177, 178, 179) is, by art. 86, 



whence 



A 2 4- B 2 + C 2 — L 2 


(180) 


L = s/(A 2 + B* + C 2 ), 


(181) 


A A 


(182) 


cos. « _ £ _ V( ^ 2+ ^ 2 + C2) 


/ - B - B 


(183) 


' — L ~ s/(A 2 + B 2 + C 2 ) 


u _ C _. C 


(184) 


c0 - c - JL - V(^ 2 + b* + c*y 


Corollary. Since 




x l = 





is the equation of the plane y t 2r 1 £Ae Zoczjs of the general 
equation (143) of the first degree in space is a pla?ie, 
the perpendicular to which is inclined to the axes by 
angles^ which are determined by equations (182-184). 

130. Corollary, Since the intersection of two planes 
is a straight line, the locus of two equations of the first 
degree is a straight line, 

131. Corollary. If the equation (143) is divided by L y 
and the values (182-184) substituted in the result, it be- 
comes 

M 

COS. « X -j- COS. a' y -j- COS. «" Z = yr, 



§ 133.] LINEAR LOCUS. 101 

Angle of planes. 

which, compared with (1 28), leads to the conclusion that 

M 

is the length of the perpendicular let fall upon the plane 
from the origin, 

132. Problem. To find the angle of two planes. 
Solution. Let their equations be 

A-x + By + Cz-\-M = 
A l x + B 1 y + C 1 z + M 1 =0, 

and let a p Y , a x p x y x be the angles which the perpendiculars 
to them make with the axes of x, y,z\ and let I be the angle 
of the planes. The angle / is also that made by the perpen- 
diculars to the planes, so that, by (43), 

COS. Jz= COS. a COS. « x + COS. |5 COS. ^ -\- COS. Y COS. Y l} (185) 

and by equations (181 - 184) 

_ AA 1 +BB 1 + CC, 

COS. J = ?— ■ py 1 — ! *- 

_ AA. + BB. + CC, 

~ V(^ + jB2 + Ca)V(il?+£!+C?)' k ' 

133. Corollary. If the planes are parallel, their per- 
pendiculars are parallel, and make equal angles with 
the axes, so that 

A, A. B B t C C. 

-irj^-z-T^L^r: (187) 



102 ANALYTICAL GEOMETRY. [b.I. CH. VI. 



Perpendicular planes. 


A B C L 
A~- -Bl~ -C[- i/ 


(188) 


ind their distance apart in this case is 




M 1 M 


(189) 



134. Corollary. If the planes are perpendicular, we 
have 

/ = 90°, cos. I =z 0, 

and (185) and (186) give 

COS. a COS. a j -\- COS. |* COS. p % -f- COS. y COS. y^O (190) 

AA X + 7?JB 1 + CC X = 0. (191) 

135. Corollary. Since 

sin. /:= -v/(l — C0S - 2 I) 
we have, by (18G), 



(^L 2 +^ 2 + C 2 ) (A 2 +B 2 -\-C 2 ) 

_ ( A2 + B 2 +^)(^f+^?+^)-(^^ 1 +^^ 1 +^ 1 ) 2 
77 (^.2 _)_ij2 _j_ c 2 ) (A 2 + B 2 + C 2 ) 

iA*B 2 — < ZAA l BB 1 +A 2 BZ + A*C 2 — ZAA 1 CC 1 ) 
( 4-i4 2 C ,2 + ^2C2_2J5^ 1 CC 1 4-B 2 C 2 J 



+ A 2 C 2 -\- B 2 C 2 —2 BB 1 CC 1 -j- B 2 C 2 

(192) 



(A* + B2+C2)(A 2 + B 2 + C 2 ) 
(AB 1 —A l Bf + (AC l —A 1 Cf-\-(BC 1 —B 1 Cf 



(A 2 +B 2 + C 2 )(A 2 +B 2 + Ci) 



§ 138.] LINEAR LOCI. 103 

Angle of line and plane. 
We also have, by (181-184), 

= (COS. a COS. |? x COS. /? COS. a x ) 2 

-j- (cos. a COS. Y x COS. y COS. «j) 2 

-|- (cos. /5 cos, y 1 — COS. y COS. (Sj 2 . (193) 

136. Problem. To find the angle which a line 
makes with a plane. 

Solution. If a, ,«, 7 , are the angles which the line makes 
with the axes, and * v fo'n, those which the perpendicular 
to the plane makes, and / the angle made by the given line 
with the plane, the angle which the line makes with the per- 
pendicular to the plane will be the complement of J, and we 
shall have 

sin. /=: cos. «cos.« 1 -)-cos. /$cos. p x -f-cos. ycos. Yl (194) 

COS. 2 I =z (COS. a COS. ft COS. /S COS. «j) 2 

-f- (COS. « COS. Yl COS. y COS. « x ) 2 

-|- (cos. ?cos. Yi — cos. y cos./* J 2 . (195) 

137. Corollary. If the line is parallel to the plane, we 
have 

sin. J = = cos.« cos. a x -j- cos.;? cos.^ -(- cos.y cos.y t . ( 1 96) 

138. Corollary. If the line is perpendicular to the plane, 
we have 

« = «! , /? = ft , y = yi . 



104 ANALYTICAL GEOMETRY. [b. I. CH.VI. 

Perpendicular to plane. 

139. Problem. To find the equation of a plane, 
which passes through a given point. 

Solution. Let a , /?, y be the angles which the perpendicular 
to the plane makes with the axes, and let x', y' z' be the co- 
ordinates of the point, and p the perpendicular let fall upon 
the plane from the origin. 

The equation of the plane is 

COS. a.Z-\- COS. /S . y -f- COS. y . Z = p, 

and since the point is in this plane, its coordinates satisfy the 
equations of the plane, and we have 

cos. « . x' -|- cos. p . y' -j- cos. y • z ' == P > 
and if the value of p thus obtained is substituted in the equa- 
tion of the plane, it gives 

cos.«(x — x') + cos.?(y — y) + cos. y {z — z') = 0, (197) 
in which «, p, y are arbitrary. 

140. Corollary. The distance of this plane from another plane 
parallel to it, and which passes at the distance pi from the 
origin, is 

p — pi = cos.a.x'-f- cos. /» . y' -\- cos. y . z' — p v (198) 

which is therefore the length of the perpendicular let 
fall from the point x' , y', z', upon the pla?ie, whose 
equation is 

cos a . % -\- cos. £ . y -j- cos. y . z =1 p x . 

141. Examples involving Linear Loci. 

1. To find the locus of all the points so situated in a plane, 
that m times the distance of either of them from a given line, 



§ 141.] LINEAR LOCUS. 105 

Examples of linetir loci. 

added to n times its distance from another given line, is equal 
to a given length. 

Solution. Let the first given line be the axis of x, and let 
the intersection of the two lines be the origin, and « the angle 
which these lines make with each other. Then, if x, y are 
the coordinates of one of the points of the locus, we have 

y = the distance from the first line, 

and if p is the distance from the second line, we have, by 
(170), 

p z= x sin. a — y cos. «. 

If, then, I is the given length, 
l—my-{-np Q 

I = m y -\- nx sin. « — n y cos. « 
— n . sin. a .x -\- (n cos. « — m) y z= — I 

so that the required locus is, by art. 108, a straight line, in- 
clined to the first given line by the angle (*, such that 



tang. p =z 



n cos. a — m 

and which, by art. 109, passes at a distance from the intersec- 
tion of the two lines equal to 

I I 

\/[?i 2 sin.' 2 «-J-(rtcos.« — m)' 2 ] \/(n 2 -\-m 2 — 2y/JKcos.«)' 

Scholium. The whole length of the line thus obtained 
satisfies the algebraical conditions of the problem, but not the 
intended conditions. For at those points, where the value of 
y or that of p Q is negative, / is no longer the absolute sum of 
my, and np but their difference. 



106 ANALYTICAL GEOMETRY. [s/l. CH. VI. 

Examples of linear loci. 

Corollary. When m — n 
we have 

cos.« — 1 — 2sin. 2 Aa 

cotan. s — — ; = 7T-. — z. —zr~ = — tang. \ « 

sin. a 2 sin. £ a COS. ^ a 

(5=90°+ J a, 

and the distance from the point of intersection becomes 
I I 



w\/(2 — 2 cos. a) " 2 m sin. J a" 

2. To find the locus of all the points so situated in a plane, 
that the difference of the squares of the distances of either of 
them from two given points in that plane is equal to a given 
surface. 

Ans. If 2 a = the distance of the two given points apart, 
and if the given surface is a parallelogram, whose base is a and 
altitude b, the required locus is a straight line, drawn perpen- 
dicular to the line joining the given points, and at a distance 
equal to £ b from the middle of this line. 

3. To find the locus of all the points, from either of which 
if perpendiculars are let fall upon given planes, and if the 
first of these perpendiculars is multiplied by m, the second by 
m lt the third by m 2 , &c, the sum of the products is a given 
length /. 

Ans. If 

cos. « a ; -j- cos. p y -f cos. y z — p 
cos. «i x + c° s - Ciy + cos. /i z — : pi 
&c, are the given planes, the required locus is a plane, whose 
equation is 

(»»cos.a-(-»i 1 cos.a 1 +&c.)x+(mcos.iS-fOT 1 cos./» 1 -f-&c.)y 

-[-(mcos.y-J-WjCOS.yj -j- &c.)z — 7-f mp-j- m iPi + &c. 






«§> 141.] LINEAR LOCUS. 107 

Examples of linear loci. 

or if the letter S. is used to denote the sum of all quantities 
of the same kind, so that 

S . m = m -f m 1 -f &c. 

the equation of this plane ma} r be written 

8. m cos. o. x -f S .m cos. P .y -}- S . m cos. 7 .z — l-\-S. mp. 

Scholium. This result is subject to limitations, precisely 
similar to those of example 1. 

4. To find the locus of all the points, whose distances from 
several given points is such that if the square of the distance 
of either of them from the first given point is multiplied by 
m 1 , that of its distance from the second given point by m 0} 
&,c. the sum of the products is a given surface V. The 
quantities m 19 ?n 2 , &c. are some of them to be negative, and 
subject to the limitation that their sum is zero. 

Ans. If x x , y x , z t is the first given point, x 2t y 2 , z 2 the 
second one, &-c, and if S is used as in the preceding ex- 
ample, we have 

S. *! = 0, 
and the required locus is the plane whose equation is 

xS.m 1 T 1 .+yS.m 1 ij l .+zS.m x z 1 =S.m 1 (x\+y2J r z2).--V 



108 ANALYTICAL GEOMETRY. [b. I. CH. VII. 

Reduction of the equation of the second degree in a plane. 

CHAPTER VII. 

EQUATION OF THE SECOND DEGREE. 



142. The general form of the equation of the second 
degree in a plane is 

Ax^ + Bxy + Cy 2 + Dx + Ey + M.=.0, (199) 

and that of the equation in space is 

Ax* + Bxy + Cy* + Dxz + Eyz -j- F z* 

-r Hx + Iy+Kz + M — 0. (200) 

143. Problem. To reduce the general equation of 
the second degree in a plane to its simplest form. 

Solution. I. By substituting in (199) equations (18) and 
(19) for transformation from one system of rectangular co- 
ordinates to another, the origin being the same ; representing 
the coefficients of xf, y\, x l} and y Y by A 19 B 1 ,D 1 , and Ej ; 
and taking a of such a value that the coefficient of x\ y\ may 
be zero; (199) becomes 

Aix\ + Biy\ +2> J x 1 + £ 1 y 1 -f 31=0. (201) 

in which we have 

A 1= zA cos. 2 « -f B sin. « cos. « + Csin. 2 « (202) 

B ! = A sin. 2 <* — B sin. « cos. « -f C cos. 2 a (203) 

DiznD cos. « -}- E sin a (204) 

Ei =—D sin. « + E cos. a, (205) 



§ 144.] QUADRATIC LOCUS. 109 

Reduction of quadratic equation. 

and a satisfies the equation 

2 (C—A) sin. « cos. « + B (cos. 2 « — sin. 2 «) =. 0. (206) 

II. If, now, we substitute the formulas (20) for transposing 
the origin in (201) ; using x 2 and y 2 for the new coordinates ; 
take the coordinates a and 6 of the new origin of such values, 
that the coefficients of x 2 and y 2 may be zero ; and denote 
the sum of the terms which do not contain x 2 or y 2 by M x ; 

(201) becomes 

A 1 x*+B 1 y2+3f 1 = } (207) 
in which 

M 1 = A 1 a* + B 1 b* + D 1 a + E x b + M, (208) 
and a and b satisfy the equations 

2A 1 a + D 1 = (209) 

2B t b + E 1 = Q. (210) 

The form (207) to which the given equation is thus re- 
duced is its simplest form. 

144. Corollary. If we take L, L' such that 

L — 2 A cos. « + B sin. a (211) 

L—2C sin. a + B cos. «, (212) 

these values may be substituted in (206), and the double of 

(202) would give 

2 A t = L cos. a -j- L' sin. « (213) 

== L' cos. « — L sin. «. (214) 

The product of (213) by cos. «, diminished by that of 
(214) by sin. «, reduced by means of the equation 

sin.2« + cos2«— 1 (215) 

10 



110 ANALYTICAL GEOMETRY. [b.I. CH.YII. 

Reduction of quadratic equation. 

is, by (211), 

2 A j cos. ct—L = 2A cos. « + B sin. «, (216) 

or 2(^ 1 — ^l)cos.«— .,Bsin.« — 0. (217) 

The product of (213) by sin. « added to that of (214) by 
cos. « is, by (215) and (212), 

2A ± sin. «z=I / = 2 Csin. a + B cos. « (218) 

2 (.4 t — C) sin. « — B cos. « = 0. (219) 

The product of (217) by 2(A 1 — C) added to that of 
(219) by B is, when divided by cos. «, 

4:{A 1 — A) {A 1 — C) — B2 — t (220) 

from which equation the value of A x may be determin- 
ed, that is, if we put X instead of A X) A x is a root of 
the quadratic equation 

4(JT— ii) (X— C) — £ 2 = 0; (221) 

the roots of which are 

X=%(A + C)±iV(B2 + A2 — 2AC+C2) 

= J(^ + C)±i\/[S 2 + (^ — C) 2 ]. (222) 

145. Corollary. If we take ij and Z/j such that 

Z x = 2 ^4 sin. « — B cos. a (223) 

&i = 2 C cos. « — B sin. a, (224) 

these values may be substituted in (206) and the double of 
(203), and give 

2 B x ~ L x sin. a + L> x cos. « (225) 

= L\ sin. « — L x cos. «. (226) 



§ 146.] QUADRATIC LOCUS. Ill 

Inclination of axis. 

The product of (225) by sin. a diminished by that of 
(226) by cos. « is, by (215) and (223), 

2B 1 sin.« = L L — 2 A sin. « — B cos. a (227) 

or 2 (B 1 — .4) sin. « + ^ cos. « = 0. (228) 

The product of (225) by cos. a added to that of (226) by 
sin. a is, by (215) and (224), 

2jB x cos. « = L\ — 2 Ccos. « — ^ sin. « (229) 

or 2(B 1 — C) cos. « + Bsin. « — 0. (230) 

The product of (228) by 2 (jB x — C) diminished by that 
of (230) by B, and divided by sin. «, is 

4 (B 1 — A) (B x — C) — B2 = 0, (231) 

from which equation the value of B 1 may be determined; 
that is, if we put X instead of B 1} B x is a root of the equa- 
tion (221). 

A x and B x are then the two roots (222) of the equa- 
tion (221). 

146. Corollary. The value of « may be obtained from the 
equation (217), which gives 

sin. « 2(ii. — A) y«rtrti 

tang, a — = v * \ (232) 

6 cos.« B v } 

or it may be obtained directly from (206). 

If we substitute in (206) 

sin. (2 a) = 2 sin. a cos. «, cos. 2 « — cos. 2 « — sin. 2 «, (233) 
it becomes 

(C— 4) sin. 2« + £cos.2<* = 0; (234) 



112 ANALYTICAL GEOMETRY. [b. I. CH. VII. 

Case of two lines. 



whence 



sin 2 K JB 

tang. 2 « = -^- = „ ^ (235) 

6 cos. 2 a A—C K ' 



147. Scholium. The values of ^ and # x (222) are al- 
ways real as well as that of a (235), and those of D x and E x 
(204) and (205) ; but the equations (209) (210) are impossi- 
ble if A x and B x are both zero, while D x and E x are not 
zero, or if either A x or B x is zero, while the corresponding 
value D x or E x is not zero. 

148. Scholium. The values of ^4 X and .Bj cannot both be 
zero, for, in this case, the quadratic terms would disappear 
from (201), and (201) could not, then, by art. 100, be a re- 
duced form of a quadratic equation. 

149. Scholium. If either A x or B x were zero, the corre- 
sponding root of (221) would be zero; that is, this equation 
would be satisfied by the value 

X=0, 

which reduces it to 

4 AC— £2 = 0; (236) 

and if we take A x for the root which vanishes, we have, 
by (232), 

tang.« = -^. (237) 



But 



COS. a - J_ — } — ; (238) 

sec.« /v/(l + tan - a ) 



§ 150.] QUADRATIC LOCUS. 113 

Case of two lines. 

whence 

sin. » = cos. a tang. « = — ^^ + — ^ (240) 

so that Z> x will vanish, only when 

DB =2AE ; (242) 

and in this case (201) becomes 

B iy \+E iyi +M=Q; (243) 

which gives 

yi = *,W( S -4iM/) ; (344) 

so that the required locus is the combination of two lines 
drawn parallel to the axis of x x at the distances from 
it equal to these two values of y lf unless these values 
are imaginary or equal, in the former of which cases 
there is no locus, and in the latter the given equation 
is the square of the equation of the line. 

150. Scholium. If the values of A, B, C satisfy (236), so 
that one of the roots of (221) is zero, and if this one is taken 
for A xi we have for the other root, by (222), 

B 1 =i(A + C)+W(^^C+A^-2AC+C^) 

= i(A + C)+i(A + C) = A + C, (245) 

10* 



114 ANALYTICAL GEOMETRY. [b. I. CH.VII. 

Parabola. 

and (201) becomes 

(A + C)y\ + D 1 x 1 +E l2/l +M = 0. (246) 

The origin may now be transposed as in art. 143, the co- 
ordinates a and b being taken of such values that the coefficient 
of y 2 may be zero, and the sum of the terms which do not 
contain % 2 or y 2 may be zero, and (246) is thus reduced to 

(A + C)y% + D 1 x 2 =,0. (247) 



The values of a and b satisfy the equations 






2(A + C) b + E 1 — 




(248) 


(A + C)b 2 + D 1 a+E 1 b + M 


= 0, 


(249) 


whence 






- 2(4 + C) 




— (A-\-C)b 2 —E 1 b — 


M 




and if we put 






*P~ A + C y 




(250) 


(247) becomes 






y% — *P X 2 = °> 






or yf = 4pz 2 . 




(251) 



151. Corollary. If the equation (221) is written in the form 

S 2 — [A + C) 8+ i (4 AC— B 2 ) = 0. (252) 

The term £(4 AC — B 2 ) is the product of the roots A , 
and JBj of this equation. 

A x and B x are therefore of the same sign, when 



§ 154.] QUADRATIC LOCUS. 115 

Ellipse. Point. 

A AC is greater than B 2 ; and they are of opposite 
signs if A A C is less than B 2 . 

152. Corollary. When B 2 is less than 4 A C, and, conse- 
quently, A x and B x are of the same sign, we will put 
A x 1 _Bj_ _1_ 

that sign being prefixed to M lt which renders the first mem- 
bers of these equations positive. If then (207) is divided by 
4- M x , the quotient is 



■^±1 = 0. (254) 



153. Scholium. If M x were zero, the equations (253) would 
be absurd, but in this case equation (207) would be 

A i x2 2+B 1 t/ 2 = (255) 

in which both the terms of the first member have the sign, so 
that the equation can only be satisfied by the conditions 

x 2 == 0, y 2 — 0, (256) 

which represents the origin of the axes of x 2 and y 2 . 

Hence, and by art. 143, the locus of the given equation 
is , in this case, the point whose coordinates are the 
values of a and b (209) and (210). 

154. Scholium. If M x were of the same sign with A x and 
B lt the upper sign would be used in equations (253) and 
(254), the first member of (254) would then be the sum of 
three positive quantities, and could not be equal to zero. 

The given equation has, then, no locus, in this case. 



116 ANALYTICAL GEOMETRY. [b. I. CH. VII. 

Hyperbola. Two lines. 

155. Scholium. When M x is of the sign opposite to that 
of A x and B x , the lower sign must be used in equations 
(253) and (254), and (254) becomes, by transposition and 
omitting the numbers below the letters, which are no longer 
necessary, 

J £r + J §s= 1 > (257) 

which is of the same form with the equation (69) of the 
ellipse. 

156. Corollary. When B 2 is greater than 4 AC, and, con- 
sequently, A x and B x are of opposite signs, we will put 

^7 = ^1' wk^^f (258) 

those signs being prefixed to M x , which render the first mem- 
bers of these equations positive. If, then, (207) is divided by 
-4- M x , the quotient is 

157. Scholium. If M x were zero, the equations (258) could 
not be used, but in this case equation (207) would be 

A x z 2 + B x y 2 = 0, 

which, multiplied by A x , gives 

A 2 x 2 = -A x B x y 2 ; 

or, extracting the root, 

A x z 2 = ±*/(-A x B x )y 2 ;^ (260) 

the second member of which is real, because A x and B x are 
of-opposite signs. 



§ 159.] QUADRATIC LOCUS. 117 

Hyperbola. Ellipse. 

The locus of the given equation is then the combi- 
nation of the two straight lines represented by the two 
equations included in (260), each of which passes 
through the origin of x 2 and y 2 . 

153. Scholium. If iT7 L is not zero, equation (259) may, by 
omitting the numbers below the letters and transposing the 
terms, be written in one of the forms 

^ 2 -_ll = i (261) 

A* B 2 \ ' 

q/2 j-2 

and the second of these equations becomes the same as the 
first by changing x, y, A, B into y, x, B, A respectively. 

Equation (261) is of the same form with equation 
(85) of the hyperbola, 

159. Theorem. The equation (257) is necessarily 
that of an ellipse. 

Proof To prove this it is only necessary to show that each 
point of its locus is so situated, that the sum of its distances 
from two fixed points is always of the same length. By com- 
paring the equation (226) with the solution of example 2, art. 
98, it is apparent that, since all the points of the ellipse satisfy 
the equation (69), they are in the required locus ; so that if 
all the points of- the required locus are in the ellipse, the two 
fixed points must be in the axis of x at a distance c from the 
origin such that 

c = ± V(A 2 — B2), (263) 

and that the given length must be 2 A. 



118 ANALYTICAL GEOMETRY. [b. I. CH. VII. 

Ellipse. 

Now the distance r of the point x,y from one of these fixed 
points is, by art. 23, 

r = V[(*-<) 2 +3/ 2 ]; (264) 

B 2 x 2 
but since y 2 ~B 2 ~- and c 2 — A 2 — B 2 , 

we have 

r == \/(x 2 — 2cx + c 2 +y 2 ) 

(B 2 x 2 \ 

(A* — B 2 \ 

Az Z2-2CZ + A2) 

-A- A ) = ±-A— W* 

Now of the two signs -|- and — , that must be used which 
gives the distance r positive. But we have 

c < A and x < A 

for c=.*/(A 2 —B 2 ) 

and z^V^A 2 --^}. 

Hence 

cx<A 2 or cs — A 2 <0; (266) 

so that the lower sign must be used in (265), which gives 

r = A- C -£; (267) 



§ 160.] QUADRATIC LOCUS. 119 

Hyperbola. 
so that for the distance from one of the fixed points we have 

r^A-l^^l, (26 8) 

and for the distance from the other 

whence r, + r 2 — 2 A ; (270) 

that is, all the points of the required locus belong to the 
ellipse. 

160. Theorem. The equation (261) is necessarily 
that of an hyperbola. 

Proof. The proof is the same as in the preceding theorem, 
except that the word difference is to be used for sum, the siom 
of B 2 is to be changed, and in the value of r (265) the lower 
sign is to be used, where c and x are both positive or both 
negative. For, since the values of c and x are 

c = ± V^ 2 + # 2 ) and^± (a* + ii|!\ 
we have, when c and x are of the same sicrn 

c x > A 2 or ex — A 2 > ; 

whence r% = f| — A. (271) 

But if c and x have opposite signs the product c x is nega- 
tive, so that 

r 2 = C ~J + A (272) 



120 ANALYTICAL GEOMETRY. [b. I. CH.VII. 

Parabola, 
whence r 2 — r 1 = ZA; (273) 

that is, all the- points of the required locus belong to the 
hyperbola. 

161. Theorem, The equation (251) is necessarily 
that of a parabola. 

Proof. We have only to show that the distance of each 
point of the locus from that point of the axis of x 2 , whose 
distance from the origin is p, is equal to its distance from 
that line which is drawn parallel to the axis of y 2 , and at 
the distance — p from it. Now since the distance of the 
point x,y from the axis of y is x, its distance from the line 
parallel to it must be 

x+p; 

and its distance r from the fixed point must be 

r = V[(*— p) 2 +y 2 ] 

— \/(x 2 — 2px+p 2 +4pz)=*/(z 2 + 2 j pz-f-;> 2 ) 

= x+p, (274) 

which is the same as the distance from the line ; all the points 
of the locus of equation (251) are then upon the same pa- 
rabola. 

162. Theorem. In different ellipses which have the 
same transverse axis, the ordinates which correspond to 
the same abscissa are proportional to the conjugate 
axes. 

Proof. Let the common transverse axis be 2 A, the differ- 
ent conjugate axes 2 5, 25j, &c, and let the ordinates, 






§ 166.] QUADRATIC LOCUS. 121 

Ratio of ordinates in ellipses and circles. 

which correspond to the same abscissa x, bey, y x &c, we 
have 

A 2 y 2 — B 2 (Al — x 2 ) 
A 2 y\ — B 2 {A 2 —x 2 ), 
whence, by division, 

A 2 y 2 : A 2 y 2 — B 2 (A 2 —x 2 ) : B 2 (A 2 —x 2 ) 
or y 2 : y 2 = B 2 : B 2 , 

or extracting the square root 

y:y x = B:B 1 = 2B:^B 1 . 

163. Corollary. Since the ellipse, whose conjugate 
axis is equal to its transverse axis, is a circle, the ordinate 
of an ellipse is to the corresponding ordinate of the 
circle, described upon the transverse axis as a diameter, 
as the conjugate axis is to the transverse axis. 

164. Corollary. In different ellipses which have the 
same conjugate axis, the abscissas which correspond to 
the same ordinate are proportional to the transverse 
axes. 

165. Corollary. The abscissa of an ellipse is to the 
corresponding abscissa of the circle, described upon the 
conjugate axis as a diameter, as the transverse axis is 
to the conjugate axis. 

166. Corollary. It may be proved in the same way 
that in different hyperbolas, which have the same trans- 
verse axis, the ordinates which correspond to the same 
abscissa are proportional to the conjugate axes ; and 

11 



122 ANALYTICAL GEOMETRY. [b. I. CH.VII. 



Ratio of ordinates in hyperbolas. 



that in different hyperbolas, which have the same con- 
jugate axis, the abscissas, which correspond to the same 
ordinate, are proportional to the transverse axes. 

167. Corollary. Understanding, by an equilateral 
hyperbola, one in which the axes are equal, the ordi- 
nate of any hyperbola is to the corresponding ordinate 
of the equilateral hyperbola, described upon its trans- 
verse axis, as the conjugate axis is to the transverse 
axis, and the abscissa of the hyperbola is to the cor- 
responding abscissa of the equilateral hyperbola, de- 
scribed upon its conjugate axis, as the transverse axis 
is to the conjugate axis. 

• 

168. The term abscissa is often applied, in regard to 
the ellipse and hyperbola, to denote the distance of the 
foot of the ordinate from either of the extremities of 
the transverse axis. 

Thus the abscissas of the point M (fig. 38.) of the ellipse 

are 

CP = AC — AP — A — x 

and CP = AC + AP = A + x. 

The abscissas of the point if (fig. 36.) of the hyperbola are 

CP = AP — AC z=% — A 

and CP ~ AP + AC— x + A. 

169. Theorem. The squares of the ordinates in an 
ellipse or hyperbola are proportional to the products of 
the corresponding abscissas, the term abscissa being 
used in the sense of the preceding article. 



§ 170.] QUADRATIC LOCUS. 123 

Ratio of ordinates in ellipse and hyperbola. 

Proof. I. The product of the abscissas for the point 
x, y of the ellipse is, by the preceding article, 

(A + x) {A — x) — A 2 — x 2 ; 
and this product for the point x' f y' is 
A 2 — x' 2 . 
But, by equation (68), we have 

A 2 y 2 =z A 2 B 2 —B 2 _x 2 
A^y'^— A 2 B 2 —B^x'S; 
whence 

A 2 y 2 : A 2 y' 2 = A 2 B 2 — B 2 x 2 : A 2 B 2 — B 2 x' 2 , 
or, reducing to lower terms, 

y 2 : y' 2 — A 2 — x 2 : A 2 — x' 2 y 
which is the proposition to be proved. 

II. In the same way, for the hyperbola, the products of the 
abscissas for the points x, y, and x', y' are 

X 2 _ A 2 and x> 2 — A 2 . 
But, by equation (84), 

A 2 y 2 — B 2 x 2 — A 2 B 2 
A2y<2 — b 2 x' 2 — A 2 B 2 , 
whence y 2 : y' 2 = x 2 — A 2 : x' 2 — A 2 . 

170. Theorem. The squares of the ordinates in a 
parabola are proportional to the corresponding abscis- 
sas. 



124 ANALYTICAL GEOMETRY. [B.I. CH.VII. 

Angle inscribed in a semiellipse. 

Proof. For the point x, y we have by (93) 
y* = 4Px, 
and for x',y' y' 2 = 4P x', 

whence y 2 :y 2 = 4Px:4 Px' = x : x', 

which is the proposition to be proved. 

171. Problem. To find the magnitude of an angle, 
which is inscribed in a semiellipse. 

Solution. Let CMC (fig. 45.) be the semiellipse, whose 
semiaxes are A and B, let I be the required angle CMC\ 
a the angle MCX, £ the angle MCX i and x' t y' the co- 
ordinates of the point M. 

Because the line MC passes through the point x', y' and the 
point C, whose coordinates are 

y = 0, x — ACz=A, 

we have, by art 121, 

tang -" = F=^i ; < 275 > 

and, because the line MC passes through the point x' y' and 
the point C", whose coordinates are 

y = 0, x — — 4, 
we have 

tan g-P = jqjTZ ; (276) 



hence 



tang. I = tang. (/* — «) = i?^i l^BiLL 

s 6 v ; 1 + tang. <* tang. /s 

2 Ay' 



% l2—.A2+y<2 



§ 174.] QUADRATIC LOCUS. 125 

Supplementary chords. 



But, by (68), 



and, therefore, 



A 2 



ZAB*yi 2AB 2 , rtl ^ 



172. Corollary. The product of (275) and (276) gives by 
the substitution of 

B 2 

y>* =± - 2 (A* - **) 

tang, a . tang. ^ = — — , (278) 

which is the condition that must be satisfied by the two angles 
a and p, in order that two lines CM and CM 1 , drawn from 
the two points M and M', may meet upon the curve. 

Two such lines are called supplementary chords; 
so that (278) is the condition which expresses that two 
chords are supplementally. 

173. Corollary. If equation (278) is compared with (72), 
it is found to be identical with it ; so that the condition that 
two chords are supplementary is identical with the condition 
that two diameters are conjugate. 

If then a given chord, as CM, is parallel to a given 
diameter B 1 AB 1 , the chord CM, supplementary to 
CM, is parallel to the diameter C 1 AC' 1} conjugate to 
B X AB\. 

174. Problem. To draw a diameter, which is conju- 
gate to a given diameter. 

11* 



126 ANALYTICAL GEOMETRY. [b. I. CH. VII. 

Chords bisected by diameter. 

Solution. Let B 1 AB\ (fig. 45.) be the given diameter. 
Through C draw the chord CM parallel to B 1 AB\\ join 
CM and the diameter C^C'^ which is drawn parallel to 
CM, is, by the preceding article, the required diameter. 

175. Problem. To find the magnitude of the angle 
formed by two chords drawn from a point of the hyper- 
bola to the extremities of its transverse axis, which are 
called supplementary chords. 

Solution. The solution is the same as that of art. 177, ex- 
cept in regard to the sign of J5 2 , which being changed gives 
for the required angle I 

2AB2 ' 

176. Corollary. The corollaries of arts. 172, 173, 
and the construction of art. 174, may then be applied 
to the hyperbola, and equation (88) is the condition 
that two chords are supplementary. 

177. Theorem. The chords which are drawn parallel 
to the conjugate of any diameter of an ellipse or hyper- 
bola are bisected by it. 

Proof. For each value of x there are two equal values of y, 
one positive the other negative, which are, in the ellipse, 



y = ±^V(4 2 -* 2 ) 



and, in the hyperbola, 



j = ±^V(« ! -i ! ); 



<§> 180.] QUADRATIC LOCUS. 127 

Parameter. 

so that if for the value of x equal to AP (fig. 46.), the line 
MPM' is drawn parallel to the conjugate diameter, and if 
PM, P M' are taken each equal to the absolute value of y, 
the points 31, M' are upon the curve, and the chord MM', 
which joins these points, is bisected at P. 



178. Corollary. The same proposition and proof may 
be applied to the parabola, using the word axis instead 
of diameter. 

179. Corollary. The chords drawn perpendicular to 
either axis of an ellipse or hyperbola, or to the trans- 
verse axis of the parabola are bisected by this axis. 

180. Problem. To find the length of the chord 
drawn through the focus of the ellipse, the hyperbola 
or the parabola, perpendicular to the transverse axis ; 
this chord is called the parameter of the curve. 

Solution. I. Represent the parameter of the ellipse by 
4p; and its half or the ordinate is 2p, the corresponding 
abscissa being, by example 3, art. 98, 

c = */{A 2 — B 2 ) or c 2 — A 2 — B 2 . 

Hence the equation of the ellipse gives 

2Ap — B \/(A 2 — c 2 ) = B 2 

II. In the same way in the hyperbola we should find the 
same values of 2jp and 4 p. 



128 ANALYTICAL GEOMETRY. [b. I. CH. VII. 

Tangent. 

III. In the parabola whose equation is 
y 2 s== 4 p x 
the abscissa for the parameter is p ; at which point 
y 2 —. 4 p 2 } y ~2p 

parameter = 4p. 

181. Corollary. In the ellipse or hyperbola, we have 

A : B z= B :2p 
or 2A : 25 = 25 : 4p; 

so that £Ae parameter is a third proportional to the 
transverse and conjugate axes. 

182. Theorem. The line draxon through either ex- 
tremity of a diameter of the ellipse or hyperbola, parallel 
to the conjugate diameter, is a tangent to the curve. 

Proof For the two values of y are equal to zero at 
the point, so that either of these lines has only one 
point in common with the curve. 

183. Problem. To draw a tangent to the ellipse or 
hyperbola at a given point of the curve. 

Solution. Join the given point M to the centre A. Through 
the extremity C of the transverse axis draw the chord CM 1 
parallel to AM. Join CM, and the line drawn through the 
parallel to CM is, by arts. 179 and 175, the required tan- 
gent. 

184. Scholium. The drawing of tangents to these 
curves will be more fully treated of in a subsequent 
chapter. 



§ 185.] QUADRATIC LOCUS. 129 

Reduction of quadratic equation in space. 

185. Problem. To reduce the general equation of the 
second degree in space to its simplest form. 

Solution. I. Substitute the equations (40, 41, 42) in (200), 
making 

fl = 0, 6 = 0, c = ; 

so that the direction of the axes may be changed without 
changing the origin. 

If we represent the coefficients of x\, y\ y z\, x lt y lt z x by 

ij=i COS. 2 a + B COS. « COS. a ' -f- C COS. 2 «' 

-f- D COS. « COS. a" -\- E COS. a' COS. a" -f- -FcOS. 2 a" 

B 1= A cos. 2 /? + B cos. p cos. /s' + Ccos. 2 ^ 

+ D cos. |S cos. /s" + JB cos. /s' cos. /?" + JFcos. 2 /J" 

C x z=z A cos. 2 y -\- B cos. y cos. y' ~f- C cos. 2 y' 

+ D cos. y cos. y " -\- E cos. y' cos. y" -f- -Fcos. 2 y" 

jfiT x = ii" COS. a -f- I COS. «' + X COS. a" 

I x —H cos. /9 + J cos. F + K cos. /J" 
K x = H cos. y -}- 7 cos. y' + -K" cos. y" 

and take a, <s } y, a', ,*', y', a ", /?", y " to reduce the coefficients 
°f x i V \> x \ z \* V \ z \ t0 zero i tnat i s > t0 satisfy the equations 

= 2 A COS. « COS. j5 + 2 CCOS. a! COS. /*' + 2 FCOS. a" COS. |S" 

4-jB(c0S.aC0S. 1 ^ / -(-C0S.a / C0S. 1 '5)-|-Z)(c0S.«COS.^ // -|-C0S.« // C0S.^) 
+ E (cos. a' COS. »" + COS. a" COS. ?') (280) 

= 2 J. cos. a cos. y + 2 Ccos. a' cos. y' + 2 i^cos. «" cos. y" 

4~-B(cos.acos./-(-cos.« / cos.y)-|— D(cos.acos.y // -f-cos.« // cos.y) 
+ E (cos. «' cos. y" + cos. «" cos. y') (281 ) 



130 ANALYTICAL GEOMETRY. [b. I. CH. VII. 

Quadratic equation in space. 
= 2 A cos. p cos. v + 2 C cos. /s' cos. y 1 -\-2E cos. |8" cos. y" 

+.B(c0S.(?C0S./-}-C0S./5 / C0S.y)-(--D(C0S.i5c0S./ / -|-C0S..' s// C0S.y) 

+ E (cos. ?' cos. y" + cos. ?" cos. /) (282 

which, combined with the six equations (44-49), completely 
determine the values of these quantities, equation (200) be- 
comes 

A^+B^+C^+H.x.+I.y.+K.z.+M^O. (283) 

II. Substitute the equations (53-55) for changing the origin 
to the axes x 2 , y 2> z 2 , and (283) becomes 

A i * 2 2+B 1 y*+C 1 zi + {ZA 1 a + H 1 )x 2 

+ (2B 1 b + I 1 )y 2 + (2C 1 c + K 1 )z 1 +3T 1 =0 > (284) 

M 1 ==A 1 f+B 1 l^+C 1 c^-H l a+I l b+K 1 c+M 9 

and in which if a, b, c are taken to satisfy the equations 

2il 1 a + JHT 1 =0 (285) 

2J5 1 6+ I x = (286) 

2 C x c + K, = (287) 

(284) becomes 

^i *§ + #i y| + ^i *i + ^ = o. (288) 

186. Corollary. If we take 

L — 2 A cos. « + 5 cos. a'+ D cos. «" (289) 

L' = 2 Ccos. «' + jB cos. « + E cos. «" (290) 

Z"= 2 .P cos. «"+ i> COS. a-\- E COS. «'. (291 ) 

These values may be substituted in (280), (281), and the 
double of the value of A lt and they give 



§ 186.] QUADRATIC LOCUS. 131 

Quadratic equation in space. 

2 A 2 = L cos. a + L cos. «' + L" cos. «" (292) 

= L cos. (S -f L' cos. |S' + L" cos. /?" (293) 

= 1 cos. y + £' cos. / + L" cos. y". (294) 

If (292) is multiplied by cos. a, (293) by cos. ?, and (294) 
by cos. y, the coefficient of L in the sum of the products is 
by (47) unity, while those of L' and L" are by (50) and (51) 
zero, so that this sum is by (289) 

2 A , cos. « = L = 2 A cos. a-\-B cos. a! -f- D cos. a", (295) 

or 2(A 1 —A)cos.a — Bcos.a' — Dcos.a" = 0. (296) 

If (292) is multiplied by cos. «', (293) by cos. s', and (294) 
by cos. /, the sum of the products is, by (48, 50, 52, 290), 

24'j cos. a' — JJ — 2 Ccos. «' + .B cos. a + E cos.«" (297) 

or 2 (4 x — C) cos. J — # cos. « — E cos. «" = 0. (298) 

If (292) is multiplied by cos. a ", (293) by cos. ?'*, and 
(294) by cos. y", the sum of the products is, by (49, 51, 52, 
291), 

2A 1 cos.c^ — L"—2Fcos.a" + Dcos, a + Ecos.a / (299) 

or 2(4 1 — F)cos.a" — j0cos.« — £cos«' — 0. (300) 

If (296) is multiplied by 4 (.4 X — C) (A 1 — F) — E* , 
(298) by 25(i 1 -P) + DE, (300) by 2Z>(^ 1 — C) 
-f- B^, the sum of the products divided by 2 cos. « is 

4(^ 1 -^)(^ 1 -C)(^ ] -^)_^(^[ 1 _^) 

— B°-(A 1 — F) — D^(A 1 — C) — BDE = (301) 

from which the value of A x may be found. 



132 ANALYTICAL GEOMETRY. [b. I. CH. VII. 

Quadratic equation in space. 

187. Corollary, Since the value of B t is obtained from 
that of A x by changing «, «', a 1 ' into P, p', p", and since by 
this same change and that of P, p', p" into y, y', y", and also by 
that of y, y', r" into «, «', «", (280) is changed into (282), and 
(281) into (280); it follows that these same changes may be 
made in the equations from (295) to (301), and (301) will 
become 

±(B 1 -A)(B 1 -C)(B 1 -F)-E*(B 1 -A) 

— B*(B 1 —F) — D*(B 1 — C) — BDE = (302) 

from which B x may be found. 

188. Corollary. Since the value of C x is obtained from 
that of jBj, by making the same changes as in the preceding 
article, and since, by these changes (282) is changed into 
(281), and (280) into (282) ; it follows that these changes 
may also be made in the equations obtained by the preceding 
article, (302) will thus become 

4( Cl -A)(C L -C) (C l -F)-&(C l -A) 

— B2(C 1 —F) — D*(C 1 — C) — BDE = (303) 
from which C x may be found. 

189. Corollary. Since the equations for determining 
A v B lt C x differ only in the letters which denote the 
unknown quantities, and since these equations are of 
the third degree, it is evident that A 1 , B lt C x are the 
three roots of the equation of the third degree 

4(X— A)(X— C)(X— F)— E*(X— A) 

— B*(X—F) — D*(X—C)—BDE=iQ. (304) 



§ 192.] QUADRATIC LOCUS. 133 



Reduction of quadratic equation. 



190. Scholium. Every equation of the third degree has at 
least one real root, so that one at least of the three quantities 
A lt B 19 C ± must be real. If we assume this one to be .4^ the 
corresponding values of cos. «, cos. «', cos. a", asdetermined by 
equations (296, 298, 300), and the 1st of art. 90, are also real ; so 
that equations (280) and (281) are satisfied without assigning 
any values to <s, /*', p" t y, /, r". If (282) is not also satisfied, let 
its second member be represented by Z> 1? and equation (200), 
instead of being reduced to the form (283), will become 

+ H 1 z 1 +I 1 y l +K l z l + 3I=0. 

If now the same transformation is effected upon this equa- 
tion, so as to transform it to the axes of x 2 , y 2 , z 2i the equa- 
tion for determining A 2 , B 2 , C 2 would be obtained from (304), 
by changing A,B, C, D, E, F into A lt 0, B 1} 0, D lt C lt 
(304) thus becomes 

4(X— A X )(X— B 1 ){X— CJ — D\(X— A 1 ) = (305) 

the roots of which are 

X=A„ 

and X=i(B 1 + C 1 )±WlD 3 1 + (B 1 -C 1 Y] (306) 

which are all real, so that the given equation can always be 
transformed to the form (283), and all the roots of (304) will 
be real. 

191. Scholium. If either A ± , B lf or C 1 is zero, one of the 
equations (285 - 287) is impossible, unless the corresponding 
value of jtJ 15 J 1? or K x is zero. 

192. Scholium. The three roots A li B x> C 1 cannot all be 

12 



134 ANALYTICAL GEOMETRY. [b.I. CH.fll. 



Cases of quadratic locus in space. 



zero at the same time ; for in this case (283) would be linear, 
and would not be a reduced form of a quadratic equation. 

193. Scholium. If A ± and H 1 are both zero, the values of 
b and c can be taken to satisfy equations (286) and (287), and 
(283) is then reduced to 

•B 1 y| + Ci^I+^ 1 = 0- (307) 

194. Scholium. If A x is zero and H x is not so, b and c 
can satisfy equations (286) and (287), and a can be taken to 
satisfy the equation 

M x p 0, 
so that (283) is then reduced to 

£ 1 y§ + C 1 *§ + ///* 1 =0. (308) 

195. Scholium. If A t and B x are zero, c can be taken to 
satisfy equation (287), and if either II 1 or I 1 is not zero, 
a or b can be taken to satisfy the equation 

M x = 0, 

so that (283) is then reduced to 

C t zl + H ll2 + J, y 2 = 0. (309) 

But if both H 1 and I x are also zero, (283) becomes 

C ± z| + M x — 0. (310) 

196. Scholium. If the values of A l , B v C lt and M x have 
all the same sign, (288) is impossible, and there is no locus. 



§ 200.] QUADRATIC LOCUS. 135 

Cases of quadratic locus in space. 

197. Corollary. If A lt B 1} C x have all the same sign, 
which is the reverse of M iy let A 2 , B 2 , C 2 be so taken, that 



(311) 



1 _A LL J_ B^ _1 C L 

A\~~~Wl B 2 ~ Ml C 2 ~~ lf~' 

and the quotient of (288), divided by — M ir ia 

-1-1+ %+l?l -*=<>■ ( 313 > 



198. Corollary. If two of the quantities A 19 B r , C x have 
the same sign with 3I lf while the other one, which we will 
assume to be A lf has the reverse sign, we will take 



1 _A_ L \ B^ _1_ C^ 

A 2 ~~ Ml B 2 ~~ Ml C%— "Ml 

and the quotient of (288), divided by — M lt is 



(313) 



<v2 ni 2 iy2 

-4-ft— ct- 1=0 - (314) 



199. Corollary. If of the quantities A lt B lf C 1} one, 
namely C x , has the same sign with M lt while the other two 
have the reverse sign, we will take 



1 


A, 1 


% 


1 


<?• 


A 2 

^2 


M t ' B%~ 


M t ' 


CI' 


~ if, 



(315) 

and the quotient of (288), divided by — M 1} is 

V 2 V 2 % 2 

A^ + B^-^- 1 ^ - < 316 ) 

■^2 ^2 ^2 

200. Corollary. The values of 2 J 2 , 2i? 2 , 2 C 2 are 



136 ANALYTICAL GEOMETRY. [b. I. CH. VII. 

Cases of quadratic locus in space. 

called the axes of the surface in either of the three last 
articles, so that the three different values of 



*<M& 



which are found from equation (304), are the semi- 
axes. 

201. Scholium. If M x is zero, the equations (311), 
(313), and (315) are impossible, but in this case (288) 
becomes 

A *l + #i 3/1 + ^*1 = 0. (317) 

202. Scholium. If A x , B xi and C x have all the same 
sign, (317) is only satisfied by the values 

x 2 = 0, y 2 = 0, z 2 =: 0, (318) 

so that the origin ofx 2 , y 2i z 2 is in this case the re- 
quired locus. 

203. Corollary. If of the three quantities, A lf JB t , C lf one, 
as C lt is negative, while the other two are positive, we will 
take 

and (317) becomes 

A % + %-i =Q - (320) 

204. The form of a surface is best investigated by- 
examining the character of its curved sections , which 



§ 205.] QUADRATIC LOCUS. 137 

Examples of quadratic loci. 

are made by different planes. The farther investiga- 
tion of the surfaces, represented by quadratic equations, 
will, therefore, be reserved for Chapter IX. 

205. Examples involving plane quadratic Loci. 

1. To find the locus of all the points in a plane, which are 
so situated with regard to given points in that plane, that jf 
the square of the distance of each point from the first given 
point is multiplied by m, the square of its distance from the 
second given point by m", &.c, the sum of the products is 
equal to a given surface V. 

Solution. Let the given points be, respectively, x', y' ; 
x", y", &,c. 

The distances of the point x, y from these points is given 
by equation (23), and we have, by the conditions of the prob- 
lem and using S, as in art. 141, 

S.m' (a; — x') 2 + S.m' (y—y'f — V, 
or 

S.m' .x* + S.m' .y* — 2 S.m 1 x' .x— 2S.m'y f .y 

+ S.m>(x / z + y<2) — V= 0. 

This equation is already of the form 201, and may be 
reduced to the form (207) by making 

S.m'x' S.m'y' 

a - ~£^T' - ~S~m T ~ 

[(S.m')* — 2]. [(£.«i'z') 9 + (#'*»'. y') 9 ] 

M * = —s^< 

+ S.m'(x' 2 +y' 2 )-V. 
12* 



138 ANALYTICAL GEOMETRY. [b. I. CH. Vlf. 

Examples of quadratic loci. 
We have then for the axes, by (253), 

so that the locus is a circle, the coordinates of whose centre 
are — a and — b, and whose radius is A 2 . 

Corollary. — M x and S.m' must be both positive or both 
negative. 

2. To find the locus of all the points in a plane, which are 
so situated with regard to given lines in the plane, that if the 
square of the distance of each point from the first given line 
is multiplied by m lt the square of its distance from the second 
line by M 2 , &c, the sum of the products is equal to a given 
surface V. 

Solution. Let the given lines be respectively 

sin. <* 1 x — cos. « 1 y = — p x 

sin. « 2 x — cos. « 2 y = — p 2> &c. 

The distances of the point x, y of the locus from these lines 
is given by equation (170), and give, by the conditions of the 
problem, and using S as before, 

S .m l (sin. « \ . x — cos « ± . y — p 1 ) 2 = Vj 

which, developed and compared with equations (199 - 256), 
give A j and B x as the roots of the equation 

4^T 2 — 4>Sf.m 1 JT+4>S , .m 1 sin. 2 « 1 >S'm 1 cos 2 a 1 ~- (jS , »« 1 sin. 2 « 1 ) 2 =0,. 

and to find «, 

_ S .m. sin. 2 «, 

tan.2« = ^ — ! —-i, 

& ,m x cos. 2 « x 

and the values of a ) b i m x may be found by equations (208-210). 



<§> 205.] QUADRATIC LOCUS. 139 

Examples of quadratic loci. 

3. To find the locus of the centres of all the circles which 
pass through a given point, and are tangent to a given line. 

Ans. A parabola of which the given point is the focus, and 
the given line the directrix. 

4. To find the locus of the centres of all the circles, which 
are tangent to two given circles. 

Ans. When the locus is entirely contained within one of the 
given circles, it is an ellipse of which the foci are the two 
given centres, and the transverse axis is the sura of the two 
given radii. Otherwise, it is an hyperbola, of which the foci 
are the two given centres, and the transverse axis the differ- 
ence of the two given radii, if the contacts are both external 
or both internal, and their sum, if one of the contacts is exter- 
nal and the other internal ; and it may be remarked, that the 
contact with either of the given circles is external upon one 
branch of the hyperbola, and internal upon the other. 



140 ANALYTICAL GEOMETRY. [b. I. CH. VIII. 

Similar ellipses. 



CHAPTER VIII. 



SIMILAR CURVES. 



206. Definition. Two curves are said to be similar 
when they can be referred to two such systems of 
rectangular coordinates, that if the abscissas are taken 
in a given ratio, the ordinates are in the same ratio. 

207. Corollary. If the given ratio is m : w, and if the co- 
ordinates of the first curve are x, y, the corresponding ones of 
the second curve must be 

nx ny 
m i m ' 

so that if these values are substituted for the coordinates in 
the equation of the second curve, the equation obtained must 
be that of the first curve. 

208. Theorem. Two ellipses or two hyperbolas are 
similar, if the ratios of their axes are equal. 

Proof. I. Let the semiaxes of the two ellipses be A, B 
and A', B 1 , we have, by hypothesis, 

A : A' .== B : B $ 

and the equations of these ellipses are (68) 

A 2 ^ B 2 ~~ 

_ x _l 4. II - 1 • 
A>2 1" b<2 — ' 



; 



§ 209.] SIMILAR CURVES. 141 

Radii vectores of similar curves. 

and if, in the second equation, we take the coordinates in the 

A'x 
ratio equal to that of the axes, that is, substitute for x, ——, 

■A. 

B'ii A'v 

and for y, — - z= — - ; it becomes identical with the first 

equation. 

II. The same reasoning may be applied to the hyperbola ; 
but it must be observed, that the ratios of the transverse axes 
must be equal to that of the conjugate axes in the two hyper- 
bolas ; and the theorem must not be applied to the case in 
which the ratio of the conjugate axis of the first curve to the 
transverse axis of the second is equal to that of the transverse 
axis of the first curve to the conjugate axis of the second 
curve. 



209. Theorem. The radii vectores, which are drawn 
in the same direction to two similar curves, are in the 
same ratio with the corresponding coordinates. 

Proof. If x and y are the coordinates for the first curve, 
and %' and y' the coordinates for the second curve, taken as 
in art. 207, we have 

x x 1 

y ~~ y 1 ' 

so that, by (11), the angle <p — «, which determines the di- 
rection of the radius vector drawn to the point x, y, is equal 
to the angle which determines the direction of the radius 
vector drawn to the point x', y'. These radii vectores must, 
therefore, coincide in direction, and we have for their values 

r — x sec. (q> — «) } r ' = x' sec. (y — «) 

jl ~ jl _ y_ 

r' x' v 1 ' 



142 ANALYTICAL GEOMETRY. [b. I. CH. VIII. 

Similar surfaces and solids. 

210. Similar surfaces may be defined in the same 
way as similar curves, and are subject to propositions 
precisely like those of arts. 207 and 209. 

Similar solids are solids bounded by similar sur- 
faces. 



$ 211.] PLANE SECTION OF SURFACES. 143 

Section of surface by a plane. 



CHAPTER IX. 

PLANE SECTIONS OF SURFACES. 

211. Problem. To find the section of a surface made 
by a plane. 

Solution. I. If the cutting plane is one of the coordinate 
planes, ihat of xy, for instance, the points of the section are 
all of them in this plane, and we have, therefore, for all these 
points 

z = 0, 

so that we have only to substitute zero for z in the 
equation of the surface to find the equation of the inter- 
section with the plane of x y. In the same way by 
putting 

x = 

the intersection with the plane of y z is found, and the 
intersection with the plane of x z is found by putting 

y = 0. 

II. For any other plane the intersection is found by trans- 
forming the coordinates of the surface, to a system of which 
the cutting plane is one of the coordinate planes. If the 
cutting plane is supposed to be the plane of x 1 y 1} we shall 
be obliged to put 



144 ANALYTICAL GEOMETRY. [b. I. CH. IX. 

Section of a surface by a plane. 

after substituting the equations (40 - 42) for the transformation 
of coordinates. But a useless operation is avoided by putting, 
at once, 

z l = 

in the equations for transformation. 

The required equation is, then, obtained by substitut- 
ing in the given equation of the surface the equations 

x — a -\- x x COS. a -\- y x COS. p (321) 

y z= b + x x cos. «' -|- y t cos. /»' (322) 

z — c-\-x 1 cos. «"+ y x cos. P". (323) 

In which a, b, c are the coordinates of a point of the 
cutting plane which is the origin of x x and y x , «, «', «", 
and P, P', P" are the angles which the two axes of x v y x 
make with the given axes. 

212. Corollary, If the cutting plane is parallel to the 
plane of x y, the axes of x 1 and y 1 may be taken parallel to 
the axes of x and y, and the origin may be taken in the axis 
of z, so that the equations (321 -323) become 

x — x lt y =zy 1} z— c. (324) 

If the cutting plane is parallel to the plane of y z, we have in 
the same way 

x — x u y =zb, z — z x \ (325) 

and if it is parallel to the plane of y z, we have 

x = a f y =zy 1} z=z z v (326) 

213. Corollary. If the cutting plane passes through the 
axis of x, the axis of x may be taken for that of x lt and the 






§ 214.] PLANE SECTION OF SURFACES. 145 

Section of quadratic surface. 

origin may remain as it was. In this case equations (321 - 
323) become 

x = x 19 y = y x cos. H', z = y x sin. p'. (327) 

If the cutting plane passes through the axis of y, and if 
the axis of y is taken for that of x 1 , (321-323) become 

x = y t cos. P, y == x lt z =: y x sin. p. (328) 

If the cutting plane passes through the axis of z, and if the 
axis of z taken for that of x lt (321 - 323) become 

x = y x cos. ?, y = y x sin. ?, z — x t . (329) 

214. Problem. To find the section of a surface of the 
second degree made by a plane. 

Solution. The equation (200) is the most general equation 
of the surface of a second degree. It may then be regarded 
as the equation of the surface referred to coordinate planes, of 
which the plane xy is the cutting plane. By putting 

z = 0, 
we have then for the required section 

Ax2+Bxy + Cyl + H x + I y + M=z 0. (330) 
From the discussion (201-262), it follows that if 

B2 — 4AC<0 
the section, if there is one, is a point or an ellipse. But if 
B 2 — AAC— 

it is a parabola, a straight line, or a combination of two paral- 
lel straight lines But if 

B 2 — 4^C>0 

it is an hyperbola, or a combination of two straight lines. 
13 



146 ANALYTICAL GEOMETRY. [b. I. CH. IX. 

Sections of quadratic surface. 

215. Corollary. For the section which is parallel to the 
plane of x y at the distance c, we have by (324) putting 

H 1 = Dc + H (331) 

I i —Ec + I (332) 

M 1 — Fc^ + Kc + M (333) 

AzZ+Bz^^CyZ+H^t+I^t+M^O; (334) 

so that this section is in the same class with that made 

by the parallel plane of x y, so far as it depends upon 

the value of B 2 — 4 AC. 

216. The values of A x and B x depend by (220, 231) only 
upon those of -4, B,, C, so that the ratios of the semiaxes 
A 2 and B 2 must also depend only upon A, B, C, and be the 
same for all the parallel sections of the quadratic locus. 

Hence, if one of the sections of a quadratic locus is 
an ellipse, all the parallel sections must be similar el- 
lipses, except those which are points. 

If one of the sections is an hyperbola, all the curved 
parallel sections are hyperbolas ; and all those sections 
are similar whose greater axes are transverse ; and also 
those are similar whose greater axes are conjugate. 

If one of the sections is a parabola, all the curved 
sections which are parallel to it are parabolas. 

In all the parallel sections the axes are parallel. 

217. Problem. To investigate the form of the sur- 
face of equation (312). 

Solution. The numbers below the letters were only used to 
distinguish the different axes of coordinates; they may, then, 
be omitted, and (312) may be written 
x 2 v 2 z 2 



$217.] PLANE SECTION OF SURFACES. 147 

Ellipsoid. 



I. The equation of the section parallel to the plane of x y 
at the distance c from the origin is 



2 c < 



-ji + ^+c*- 1 - ' (336 > 



which is impossible when 

C 2 > C 2 
it is the point 

|, = 0, yi =■ 

when c = C, 

and it is the ellipse whose semiaxes are 

:£ V(C^ - e»), ^ </{& - ^) (337) 

when c 2 < C 2 . 

II. The sections parallel to the planes of x z and y z are 
easily found in the same way, and it is evident that the sur- 
face is included by eight planes, of which two are drawn 
parallel to the plane of xy at the distances -f- C' and — C, 
two parallel to the plane of xz at the distances -\-B and — B, 
and two parallel to the plane of y z at the distances -)- A and 
— A. 

III. The section made by any other plane must then be 
limited, and must therefore be an ellipse or a point, so that 
this surface is called that of an ellipsoid, whose semiaxes are 

a, b } a 

IV. The section made by a plane passing through the axis 
of x, and inclined by an angle ?' to the axis of y is, by (327), 

x 2 /cos. 2 ,*' sin 2 *'V 

-ii+(-i^+-c^H- i = o - < 33S > 



148 ANALYTICAL GEOMETRY. [b. I. CH. IX. 

Ellipsoid. 



It is, therefore, an ellipse, whose semiaxes are A and 

/ COS.V BiD.»f \ BC 

1 ' V V-B 2 ^ C 2 f~ V(C^os.^'+B^m.^'Y K ° ' 
or if we substitute for cos. 2 P' its value, the second semiaxis is 

BC 

V[C 2 + (^ 2 — C2)sin.2/5']- ( 34 °) 

The ellipsoid may, then, be considered as generated by the 
revolution of an ellipse about the axis of x, the semiaxis of 
the ellipse, which corresponds to the axis of x, remaining con- 
stantly A, and the other axis changing from B to the value 
(339). 

The sections made by planes passing through the axes of 
y and z may be found in the same way. 

818. Corollary. If we have 

B = C 

the semiaxis (339) becomes equal to B, so that the el- 
lipse retains the same value of its second axis as well 
as of its first, during its revolution ; and the ellipsoid 
is one of revolution. The sections made by planes 
parallel to the plane of y z are, in this case, circles. 

219. Corollary. If we have 

A = B = C 

the revolving ellipse is a circle, and the surface is that 
of a sphere. 

220. Problem. To investigate the form of the sur- 
face of equation (314). 



§ 220.] PLANE SECTION OF SUFARCES. 149 

Hyperboloid. 

Solution. By omitting the numbers below the letters, (314) 
may be written 



x 2 y 2 z 2 

5C ~~ Ti* "" ~c 2 



1 = 0. (341) 



I. The section, parallel to the plane of x y, at the distance 
c from the origin is, by (253, 261), an hyperbola, of which 
the semitransverse axis, which is parallel to the axis of x, is 

^/( c 2_)_ C 2 ), and the semiconjugate is ^?V(c 2 +C 2 ).(342) 

II. The section, paralleUto the plane of xz, at the distance 
b from the origin, is an hyperbola, of which the semitransverse 
axis is parallel to the axis, and is 

A C 

^(62 i jg2\ an d t he semiconjugate is — \/(& 2 +-B 2 )- (343) 
B B 

III. The equation of the section, parallel to the plane of 
y z, at the distance a from the origin, is, by reversing the 
signs, 

1,2 2 2 n 2 

so that when a 2 < A 2 

the section is imaginary, that is, none of the surface is con- 
tained between the two planes drawn parallel to the plane of 
y z, at the distances -|- -4 and — A ; so that the surface con- 
sists of two entirely distinct branches, similar to the two 
branches of an hyperbola. 

When a 2 = A 2 

the section is reduced to the point 

y = 0, z = 0; 

when a 2 > A 2 

13* 



150 ANALYTICAL GEOMETRY. [b. I. CH. IX. 

Hyperboloid of two branches. 

the section is an ellipse, of which the two semiaxes are 

B^/^ — A*) and ^V(« 2 — ^ 2 )« (345) 

A A 

IV. The section made by any plane, which cuts both branch- 
es, is evidently an hyperbola, for no other curve of the second 
degree is composed of two branches. The section made by a 
plane, which cuts entirely across either branch without cut- 
ting the other, is an ellipse ; for this is the only curve of the 
second degree, which returns into itself, so as to enclose a space. 
The section made by a plane, which cuts one branch without 
entirely cutting across it, and without cutting the other branch, 
is a parabola ; for this is the only endless curve of the second 
degree, which consists of only a single branch. This surface 
is called that of an hyperboloid of two branches. 

V. The equation of the section, made by a plane passing 
through the axis of x and inclined, by an angle pf t to the axis 
of y is, by (327), 

It is, therefore, an hyperbola, whose semitransverse axis, di- 
rected along the axis of Xj is A, and whose semiconjugate 
axis is precisely that of (340). This hyperboloid may, then, 
be regarded as generated by the revolution of an hyperbola 
about the axis of x, the semitransverse axis remaining constant, 
and the semiconjugate axis changing in such a way, that its 
extremity describes an ellipse, whose semiaxes are B and C. 



VI. The section, made by a plane passing through the axis 
of y, and inclined by an angle (? to the axis of .t, is, by (328), 

(cos. 2 /? • sin. 9 /? 
~A^ W 



)n-3~ 1=0 - (347) 



§ 222.] PLANE SECTION OF SURFACES. 151 



Hvperboloid of two branches. 



When, therefore, 

.„ 2 a c\ n 2 

->o 



cos. 2 , ^ sin. 2 ? 
~A* C 2 



or tang. 2 ;<-p (348) 

the section is an hyperbola, of which the semitransverse axis is 

4.C 

, and the semiconiugate is B. (349) 

v/(C- 5 cos. 2 ; — ^ 2 sin.2^)' J s v ' 

C 2 
When tang. 2 § «= -p- 

the section is impossible, but every parallel section is a para- 
bola. 

C 2 
When tang. 2 s > _ 

the section is impossible, but there are parallel sections which 
are ellipses. 

In the same way and with like results, the sections may be 
found made by planes passing through the axis of z. 



221. Corollary. If we have 

B = C 

the semiaxis (340) becomes equal to B } so that the 
revolving hyperbola retains the same value of its second 
axis as at first, and the hyperboloid is one of revolution 
about the transverse axis. The sections made by planes 
parallel to the plane of y z are, in this case, circles. 

222. Problem. To investigate the form of the sur- 
face of equation (316). 



152 ANALYTICAL GEOMETRF. [B.I. CH. IX. 

Hyperboloid. 

Solution. By omitting the numbers below the letters, (316) 
may be written 

n-2 11/2 %2 

iP+h-w-" 1 = - (350) 

I. The section, made by a plane parallel to the plane of xy, 
at the distance c, is an ellipse, of which the semiaxes are 

-^ V(c* + (P) and * V(«= 2 + <"). (351) 

II. The section, made by a plane parallel to the plane of 
x z t at the distance 6, is when 

b 2 <B 2 

an hyperbola, of which the transverse semiaxis is parallel to 
the axis of x, and is 
A C 

-— */{B 2 — b2 )> and the semiconjugate is — \Z(B^-¥). (352) 

When b 2 = B 2 

the section is the combination of the two straight lines 

5=7*1? (353) 

When b 2 > B 2 

the section is an hyperbola, whose semitransverse axis is par- 
allel to the axis of z, and is 

ri A 

Z-y/fJP — B*), the semiconjugate is— \/(6 2 — J5 2 ). (354) 

In the same way and with like results, the sections may be 
found made by a plane parallel to that of y z. 



<§> 222.] PLANE SECTION OF SURFACES. 153 

Hyperboloid. 



III. The curved section made by any other plane is an 
hyperbola when it consists of two branches, an ellipse when it 
is limited, and a parabola when it consists of one infinite 
branch. 

IV. The equation of the section, made by a plane passing 
through the axis of x and inclined, by an angle Z 5 ', to the axis 
of y, is 

a+( C - O #--^>f-1=0. (355) 

When Sgj^-g^ij 

or tang.V<-^L (356) 

this section is that of an ellipse, whose semiaxes are 

A aDd V(C»cos.»/f— 5*sin.»n' (357) 

C 2 
When tang. 2 /»/ — _ (358) 

the section is reduced to the two parallel straight lines 

x 1 = dtzA 

drawn parallel to the axis of y\. Any parallel section to this 
one is a parabola. 

C 2 
When tang. 2 /*' > — (359) 

the section is an hyperbola, whose transverse semiaxis is in 
the direction of the axis of x, and is 

4, the semiconjugate is ^ {WsiaH , _ C2c0s??) (360) 



154 ANALYTICAL GEOMETRY. [b. I. CH. IX. 

Hyperboloid of revolution. 



In the same way, and with like results, the sections may be 
found, made by a plane passing through the axis of y. 

V. The equation of the section, made by a plane passing 
through the axis of s, and inclined, by an angle /s to the axis 
of x, is by (329) 

(cos. 2 /? ; sin. 2 /?\ n x 2 . , ' . Inn x 

This section is, therefore, an hyperbola, whose semiconju- 
gate axis directed according to the axis of z or x 19 is 

AB 

C, while the semitransverse is - (362) 

V{-° cos « H"-^ sm * 
AB 
or the semitransverse axis is s/[B2+{A 2_ B 2 )sin ^ ( 363 ) 

This hyperboloid may, then, be regarded as generated by the 
revolution of an hyperbola about its conjugate axis C, the 
extremity of the transverse axis describing the ellipse, whose 
semiaxes are A and B. 

223. Corollary. If we have 

A = B 

the semiaxis (363) becomes equal to A, so that the re- 
volving hyperbola retains its original axes, and the 
surface is that of an hyperboloid of revolution. The 
sections made in this case, by a plane parallel to the 
plane of x y, are circles. 

224. Problem. To investigate the form of the surface 
of equation (320). 

Solution. By omitting the numbers below the letters, (320) 
may be written 

31 +■&-■£ = «• < 364 ) 



«§> 224] PLANE SECTION OF SURFACES. 155 

Cone. 

I. The section made by a plane parallel to the plane of xy, 
at the distance c, is an ellipse, of which the semiaxes are 

Ac ■ Be /rt „_ 

-q- and — - ; (365) 

when the distance c is zero, this ellipse is a point. 

II. The section, made by a plane parallel to the plane of 
of x z, at the distance c, is an hyperbola, of which the semi- 
transverse axis, parallel to the axis of z, is 

C& 3 , . . Ab 

— - and the semiconjugate -— . (366) 

This section becomes the combination of the two straight 
lines 

Cx = ±Az, (367) 

when b is zero. 

III. The section, made by a plane parallel to the plane of 
y z, at the distance a, is an hyperbola, of which the semi- 
transverse axis, parallel to the axis of z, is 

Ca' . . Ba 

—j- and the semiconjugate — -. (368) 

This section becomes the combination of the two straight 
lines 

Cy — ^iBz, (369) 

when a is zero. 

IV. The equation of the section, made by a plane passing 
through the axis of x, and inclined by an angle p' to the axis 
of y, is 

x\ /cos. 2 ?' sin.2^\ 

T»+(-£i pr-)y?=° (370) 



156 - ANALYTICAL GEOMETRY. [b. I. CH. IX. 

Cone. 

When the condition (356) is fulfilled, this section is re- 
duced to the point 

x x = 0, y x = 0. 

But every section parallel to this one is an ellipse. 

When the condition (358) is fulfilled, the section is reduced 
to the straight line 

that is, to the axis of y 1 \ and every section parallel to this is 
a parabola. 

When the condition (359) is fulfilled, the section is the 
combination of the two straight lines 



x. ./sin. 2/?' cos.Vv ,_. 

and every section parallel to this is an hyperbola. 

In the same way, and with like results, the sections made 
by a plane passing through the axis of y may be found. 

V. The equation of the section, made by a plane passing 
through the axis of z, and inclined by an angle /s to the axis 
of x, is by (329) 

so that this section is the combination of the two straight 
lines 

x i , 7 /cos.2|S sin. 2 ?\ 

which are inclined at equal angles on opposite sides of the 



axis of x x . 



<§> 227.] PLANE SECTION OF SURFACES. 157 

Conic sections. 

This surface may then be regarded as generated by a straight 
line which passes through the origin, and revolves about the 
axis of z, inclined to this axis by a variable angle, whose tan- 
gent is 

AB m 

the surface is therefore that of a cone. 

225. Corollary. If A and B are equal, the axes 
(366) are equal, and the section parallel to x y is a 
circle: and the tangent (373) of the angle which the 
revolving lines makes with the axis of z, becomes 

A 
C" 

so that its value is constant, and the cone is a right 
cone. 

226. All the curves of the second degree may then 
be obtained by cutting a right cone by different planes, 
these curves are therefore called conic sections. 

From examining section iv. of art. 224, it appears 
that the section of a right cone is an ellipse, when the 
plane cuts completely across the cone, so as not to meet 
the cone produced above the vertex ; it is a parabola, 
when the plane is parallel to one of the extreme sides 
of the cone, so as not to meet it, nor the cone produced 
above the vertex ; it is an hyperbola, when the plane 
cuts the cone both above and below the vertex. 

227. Problem. To investigate the form of the surface 
of equation (307). 

14 



158 ANALYTICAL GEOMETRY. [b. I. CH. IX. 

Paraboloid. 

Solution. By omitting the numbers below the letters, (307) 
becomes 

By* + Cz* +31=0. (375) 

The equation of the section made by a plane parallel to the 
plane of yz is, then, the same with (375), so that the surface 
must be a cylinder, of which (375) is the equation of the 
base. 



228. Problem. To investigate the form of the sur- 
face of equation (308). 

Solution. By omitting the numbers below the letters, (308) 
becomes 

By 2 + Cz* + Hx = 0. (376) 

I. The section made by a plane parallel to the plane of y z 
is an ellipse or an hyperbola, and those made by planes parallel 
to the planes of x y and x z are parabolas. 

II. The equation of the section, made by a plane passing 
through the axis of x, and inclined by an angle ?' to the axis 
of y, is 

(B cos.2 /J/ + Csin. 2 /*') y \ + Hx x = 0, (377) 

so that it is a parabola, of which the vertex is the origin, and 
the parameter 

IT 

2i? = "" 2(5cos.2|J'+Csin.2|*')' (3?8) 

The surface may then be considered as generated by the 
revolution of a parabola, with a variable parameter, about the 
axis of Zj. It is, hence, called a paraboloid. 



<§> 230.] PLANE SECTION OF SURFACES. 159 

Cylinder. 

229. Corollary. If B and C are equal, (378) becomes 

*P = -" S ' (379) 

so that the parameter is no longer variable, and the 
paraboloid is a paraboloid of revolution. 

230. Problem. To investigate the form of the surface 
of equation (309). 

Solution. By omitting the numbers below the letters, (309) 
becomes 

Cz* + Hx + Iy=zO. (380) 

Before proceeding to investigate the sections of the surface, 
we may refer it to other axes, which have the same origin, of 
which the axis of z 1 is the same with the axis of z, and the 
plane of x 1 y 1 the same with the plane of xy. In this case, 
we have 

a = b =. c = 0, 

a n _ $1 _ y __ yl _ 90 o^ Y H __ 

(5' — «, /S — 90° + «, a' — a — 90°, 

so that (40, 41, 42) become 

x = x L cos. « — y \ sin. a (381) 

y = x x sin. « -f- # i cos » a (382) 

z = ?ll (383) 

which being substituted give, by taking a to satisfy the con- 
dition that the coefficient of y x is zero, 

Cz\ + (JEf cos. o + I sin. «) x, = 0, (384) 



160 ANALYTICAL GEOMETRY. [b. I. CH. IX. 

Two planes. 
in which « is determined by the equation 

tang. a = — . (385) 

The equation of the section, which is made by a plane par- 
allel to the plane of % x z 1} is now the same with (384) ; that 
is, the section is a parabola, and the surface is that of a cylin- 
der, of which the base is a parabola. 

231. Problem^ To investigate the form of the surface 
of equation (310). 

Solution. When (310) is possible, it is evidently the combi- 
nation of the two equations 

» a = ±V— 5 1 (386) 

each of which represents a plane parallel to the plane of a: 2 y 2 . 



DIFFERENTIAL CALCULUS 



BOOK II. 



14* 



BOOK II. 

DIFFERENTIAL CALCULUS. 

CHAPTER I. 

FUNCTIONS. 

1. A variable is a quantity, which may continually 
assume different values. 

A constant is a quantity, which constantly retains 
the same value. 

Thus the axes of an ellipse or hyperbola are constants, 
while the ordinates and abscissas are variables. 

Constants are usually denoted by the first letters of the 
alphabet, and variables by the last letters, but this notation 
cannot always be retained. 

2. When quantities are so connected together that 
changes in the values of some of them affect the values 
of the other, they are said to be functions of each other. 
Any quantity is, then, a function of all the quantities 
upon which its value depends ; but it is usual to name 
only the variables of which it is a function. 

Functions are denoted by the letters f.,f., F. } (p., V«, n "if-\ 
F.', / ., /,., &c ; thus 

/.(*), F(Z), 9 .{Z), „.{x), &C,/.'(S), &C., fA*)*M*)i fA*)> 

&/C are functions of x. 



164 DIFFERENTIAL CALCULUS. [b. II. CH. I. 

Independent variable. Construction of function. 

f.(x,y), JP.(x,y), &C. 
are functions of x and y. 

3. When variables are functions of each other, some 
of them can always be selected, to which, if particular 
values are given, the corresponding values of all the 
rest can be determined. The variables, which are thus 
selected, are called the independent variables. 

4. When a function is actually expressed in terms of 
the quantities, upon which it depends, it is called an 
explicit function. 

But when the relations only are given, upon which 
the function depends, the function is called an implicit 
function. 

Thus the roots of an equation are, before its solution, im- 
plicit functions of its coefficients ; but, after its solution, they 
are explicit functions. 

5. A function of a variable may be expressed geo- 
metrically, by regarding it as the ordinate of a curve, 
of which the variable is the abscissa. 

A function of two variables may be expressed geo- 
metrically, by regarding it as one of the coordinates of 
a surface, of which the two variables are the other two 
coordinates. 

A function is said to be constructed, geometrically, 
when the curve or surface, which expresses it, is con- 
structed. 

The inspection of the curve or surface, which thus repre- 



§ 8.] FUNCTIONS. 165 

Algebraic, logarithmic, trigonometric functions. 

sents a function, is often of great assistance in obtaining a 
clear idea of the function. 



6. Algebraic functions are those which are formed 
by addition, subtraction, multiplication, division, raising 
to given powers, whether integral or fractional, positive 
or negative. 

An integral function is one, which contains only in- 
tegral powers of the variable ; and a rational fractional 
function is a fraction, whose numerator and denomina- 
tor are both integral functions. 

Every other algebraic function is called irrational. 

Thus 

a -{- x, a — x, ax -\-by, a-\-bx-\-cx 2 -\- &c. 

are integral functions ; 

a a -\-b x-\- ex 2 

? T ~* i a'+b'x+c'x 2 

are rational fractional functions ; 

and \Zz, zj, &c. 

are irrational functions. 

7. Exponential or logarithmic functions involve va- 
riable exponents or logarithms of variables. 

Thus, a x , log. x, &c. are logarithmic or exponential func- 
tions. 

8. Trigonometric or circular functions involve trigo- 
nometric operations. 



166 DIFFERENTIAL CALCULUS. [b. II. CH. I. 

Compound, free, fixed, linear functions. 

Thus sin. x, tan. x, &c. 

are trigonometric or circular functions. 

9. Compound functions result from several succes- 
sive operations. 

Thus log. sin. x is the logarithm of the sine of x. 

10. When functions are so related, that the com- 
pound function formed from their combination is inde- 
pendent of the order in which the functional operations 
are performed, the functions are said to be relatively 
free ; otherwise they are fixed. 

Thus if the two functions <p and f are so related that the 
compound function <p.f.x is equal to the compound function 
f. <p. x, these two functions are relatively free, and this con- 
dition is algebraically 

<P.f.x=fi<p.x; (387) 

or if we omit the variable, which is often done in functional 
expressions which involve but one variable, (387) becomes 

<P.fi =fi<P. (388) 

11. A linear function is one, which leads to the same 
result, whether the operation indicated by it is perform- 
ed upon the whole of a polynomial at once, or upon the 
different terms of it successively. 

Such a function is indicated by the equation 

fi(x±y)=f.x±fiy; (389) 

and the product m x, that is, in times the variable, is a simple 
example of such a function. 



§ 13,] FUNCTIONS. 167 

Repeated functions. 

12. Theorem, The compounds of linear functions 
are linear. 

Proof. Let/ and/' be two linear functions, we are to prove 
that 

///(«±.y) = ffJx±ffJy. (390) 

Now we have from definition 

f.'{*;±y) ***/><?%/.% (391) 

and therefore 

.f.f.'(x±y) =f.(f.'x±f.>y)=f.fjx±f.f.<y (392) 
as we wished to prove. 

13. When the same operation is successively repeat- 
ed, the result is called the second, third, fyc. func- 
tion. 

Thus the log. log. x is the second logarithm of x. 
These repeated functions may be expressed by a notation 
similar to that of powers ; thus 

log. 2 x = log. log. x 

log. 3 x = log. log. 2 x = log. log. log. x, &c. 

f.*X=f.f.X 

fJX=ff.2 X= fff X) &C. 

Care must be taken not to confound f n (x) with [f(x)] n , or 
with f(x) n , which have widely different significations ; thus, 
[f(x)] n is the nth power of the function of x, f(x) n is the 
function of the wth power of x, while f n (x) is the wth func- 
tion of x. 

The common use of a different notation in the case of trigo- 
nometric functions must, however, cause them to be excepted 



168 DIFFERENTIAL CALCULUS. [b. II. CH. I. 

Zero function. 

from these remarks ; thus, sin. 71 x and sin. x n do either of them 
denote the nth power of sin. a:. Whenever we extend this 
notation to trigonometric functions, we shall indicate it by en- 
closing the exponent within brackets ; thus we shall denote the 
second, third, &c. sine of x by sin.t 2 ] x, sinPJx, &.c. 

14. By a process of reasoning, precisely similar to 
that used in the case of powers, it may be proved that 
we must have 

f. m f.n x=if. m + n x; (393) 

or, omitting the variable 

fmfn—fm + n, (394) 

This equation may be adopted as applicable to all functional 
exponents, whether positive or negative, entire or fractional ; 
and the signification of the exponent, when not positive and 
integral, must, in this case, be determined by the aid of the 
equation. 

15. Problem. To determine the signification of a 
function, of which the exponent is zero. 

Solution. Equation (393) becomes by making 

m = 0, 

f.ofnx=,f.« + *x=f.nx; (395) 

so that if we put 

f. n x — y 

we have f.°y z= y; (396) 

that is, the function whose exponent is zero is the vari- 
able itself, and this function may be represented by 
unity. 



<§> 18.] FUNCTIONS. 169 

Negative and fractional functions. 

16. Problem. To determine the signification of a 
function, of which the expoiient is negative. 

Solution. Equation (393) becomes, by making 
m — — n 

f.-"f»z=zf.°z=zz; (397) 

or if we put 

f. n x = y, 

we have 

f-*y = z; (398) 

that is, if two variables, x and y, are functions of each 
other, whatever function y may be of x, x is the corre- 
sponding negative function of y, or, as it is usually 
called, the inverse function of y. 

17. Confusion is likely to arise in the use of negative ex- 
ponents, unless it is carefully observed that many functions 
have, like roots, several different values corresponding to the 
same value of the variable. 

18. Problem. To determine the signification of a 
function, of which the exponent is fractional. 

Solution. We have, from (394) 

f m f. m f. m . . . z= ( f. m ) n — f. mn 

in which n denotes the number of repetitions of f. m . If now 

n> 

n' =z mn, m z= — 

n 

fmn—fn' 

mn n' 

fs*=f.~ T =fS 

15 



170 DIFFERENTIAL CALCULUS. [b. II. CH. I. 

Functions calculated like factors. 

that is, if m is a fraction, of which the denominator is 
n and the numerator n', the corresponding function is 
one which, repeated n times, gives the w /th repetition 
of the original function /. 



19. Theorem. When the different functions of which 
a compound function is composed are linear and rela- 
tively free, they may be combined precisely as if the 
letters which indicate them were factors instead of 
functional expressions. 

Proof. For the two equations (388) and (3S9), which apply 
to this case, are the same in form as the two fundamental 
equations of addition and multiplication, upon which all arith- 
metical and algebraical processes are founded. 

20. Corollary. The repetitions of the compound functions 

f.+A; /■+/>■ +U> &c. 
in which f f 1} f 2 > & c - are linear and relatively free, may 
be effected by means of the binomial and polynomial theorems. 
Thus 

(/ . +/i . ) n =/ .n +w/ .n-l /i . + ^^ / , l - 2/f . + &c . (399) 

(f.+f 1 -+f 2 ') n =f' n + n f' n - 1 fi- + n f- n -V 2 '+&*- (400) 

■ »(— v-vf. + fcc. 



1.2 



21. The exponents of functions are so similar to the 
exponents of powers, that they may be used in a similar 
way, and called functional logarithms ; so that if any 
function, as /., is assumed as a base, the functional log- 



$22.]' FUNCTIONS. 171 

Functional logarithms. 

arithm of any other function indicates the exponent 
which the base must have to be equivalent to this new 
function. 

Thus, if we denote functional logarithms by [f. log.], and 
if /.* = y. (401) 

we have n = [/. log.] 9. (402) 

and it may be shown, as in the theory of logarithms, that 

If. log.] <p. <p.< = [/. log.] y. + £/. log.] f .' (403) 

[/.]og.] 9 .» = ii-[/log.]v. (404) 

22. When a function has but one finite value corre- 
sponding to each value of its variable, included between 
given limits, and varies by infinitely small degrees for 
infinitely small changes in the value of its variable be- 
tween these limits, it is said to be continuous between 
these limits. 

The curve which represents a continuous function is obvi- 
ously a continuous curve. 



172 DIFFERENTIAL CALCULUS. [b. II. CH. II. 

Orders of infinitesimals. 



CHAPTER II. 



INFINITESIMALS. 



23. Theorem. Any power of an infinitesimal is infi- 
nitely smaller than any inferior power of the same 
infinitesimal. 

Demonstration. Let i be the given infinitesimal, a the ex- 
ponent of the given power, and b the exponent of an inferior 
power. We have, then, the ratio 

i a : i b z=z i a ^~ b : 1 ; 

that is, i a bears the same ratio to i b that 2**— J does to unity, 
or i a is infinitely smaller than i b . 

24. Definition. If a given infinitesimal is assumed as 
a base or standard to which all others may be referred, 
any infinitesimal is said to be of the order a, when it is 
infinitely less than any power of the base inferior to 
the a th power, and infinitely greater than any power of 
the base superior to the a th power. 



25. Corollary. If A is a finite quantity, and i the 
finitesimal which is assumed as 
be an infinitesimal of the a th order. 



infinitesimal which is assumed as the base, Ai a musi 



26. Corollary. If, in the preceding article, we make 
az= 



§ 29.] INFINITESIMALS. 173 

Negative orders of infinitesimals. 

we have 

Ai a z=z A i° z= A; 

so that a finite quantity is an infinitesimal of the order 
zero. 

27. Corollary. If, in art. 12, we make a negative, or 

a — — a 1 

we have 

Ai a = Ai— a == — = oc 
. i a 

so that infinitely great quantities may be regarded as 
infinitesimals of negative orders ; that is, an infinitely 
great quantity of the a th order is an infinitesimal of the 
— a th order. 

28. Theorem. Of two infinitesimals of different or- 
ders, that, which is of the inferior order is the infinitely 
greater. 

Demonstration. Let / and J be the two infinitesimals of 
the orders a and b respectively, a being greater than b, and 
let c be any number between a and b, and let i be the base. 
We have, by the definition of art. 24, /infinitely less than i c , 
and /infinitely greater than i c , so that lis infinitely less than 
J, agreeably to the theorem to be demonstrated. 

29. Corollary. When infinitesimals of different or- 
ders are connected together by the signs of addition or 
subtraction, all may be neglected but those which are of 
the lowest order, so that the sum or difference is of the 
same order with those of the lowest order, which are 
retained. 

15* 



174 DIFFERENTIAL CALCULUS. [b. II. CH. II. 

Order of function of infinitesimal. 

30. Corollary. The continued product of several in- 
jinitesimals is of an order equal to the sum of the orders 
of the factors. 

31. Corollary. If one or more of the factors is 
finite, the product is of the order equal to the sum of 
the orders of the other factors. 



32. Corollary. The order of the quotient of one infi- 
nitely small quantity divided by another is equal to the 
order of the dividend diminished by that of the divisor. 

33. Corollary. The order of any power of an infini- 
tesimal is equal to the order of the infinitesimal multi- 
plied by the exponent of the power ; and the order of 
any root is equal to that of the infinitesimal divided by 
the exponent of the root. 

.34. Theorem. The order of any function of an infi- 
nitesimal is equal to the product of the order of the 
infinitesimal multiplied by the order which the function 
would have if the infinitesimal were assumed as the base. 

Proof. Let I be the infinitesimal of the order a, J a. func- 
tion of 1, which would be of the order b, if I were the base, 
and let i be the base ; we are to prove that J is of the order 
ab. 

Since -J is of the order b with reference to J as a base, it is, 
by art. 24, infinitely greater than I c when c is greater than b, 
and infinitely less than I c when c is less than b\ but I c is, by 
art. 33, of the order ac, or of the same order with i ac ; and, 
therefore, J is infinitely less than i ac when a c is less than a b, 



^ 38.] INFINITESIMALS. 175 

Ratios of orders. 

and infinitely greater than i ac when a c is greater than ab; or 
J is, by art. 24, of the order a b. 

35. Corollary. The order of any function of an in- 
finitesimal of the first order is of the same order with 
the same function of the base. 

36. Corollary. The ratios between the orders of 
several infinitesimals are not changed by changing the 
base; and their orders with reference to the new base 
are obtained by dividing their original orders by the 
order of the new base referred to the original one. 

This rule cannot, however, be applied when the order of the 
new base, referred to the original one, is zero or infinity. 

37. Corollary. If I is an infinitesimal of the order a 

with reference to the base -/, i must be of the order - 

a 

with reference to / as a base. 



3S. Problem. To find the order of a \ when i is the 
infinitesimal base. 

Solution. Denote the order of a 1 by x; that of (a')" 1 is, by 
art. 33, m x ; while that of a mi is, by art. 35, x ; but we have 

(a 1 ) 771 =z a mi 
and, therefore, 

771 x = x } 
which gives 

2 = 0; 
that is, the order of a' is zero. 



176 DIFFERENTIAL CALCULUS. [b. II. CH. II. 

Infinitesimals of order zero not all finite. 

39. Problem. To find the order of log. i, when i is 
the base. 

Solution. Denote the order of log. i by x ; that of a log. i is, 
by art. 31, also x; while that of log. i a is, by art. 34, ax; but 
we have 

a log. * =: log. i a , 
and, therefore, ax = x, 

which gives x = ; 

that is, the order of log. i is zero. 

40. Corollary The order of : -. is also zero. 

log. i 

41. If i were zero, we know that a 1 , log. t, and 



log. i 

would be respectively 1, — oo and ; and their values must 
differ infinitely little from these values, when i is infinitely 

small ; that is, a 1 is finite, log. i is infinite, and - : is inn- 
log. « 

nitely small. But each of these infinitesimals has been shown 

to be of the order zero, so that there are infinitesimals of the 

order zero, which are finite, infinite, and infinitely small. 

42. Problem. To find the n th power of 1 -j- i, when 
i is infinitely small and n infinitely great, so that 

n i z= a. (405) 

Solution. The binomial theorem gives 

(l + 0^1+» i - + ^ 1 ^--+ " ( "- 1 2 ) .7 2) ' 3 + &c -(406) 

But » is infinite, and, therefore, 

n — 1 = n, n — 2 = w, &c. (407) 






$ 45.] INFINITESIMALS. 177 

Neperian logarithms. 
which substituted in (406) give, by (405), 

9j2 j2 »> 3 » 3 

= l+a + rk + UM + &c - (408) 

43. Corollary. When 

a— 1 

(408) becomes 

i l l 

(l + *y = l + l + T7 - + — - + &c. (409) 

If we denote the value of the second member of (409) by 

e, we easily find 

c = 2,71828+ (410) 

and (409) is 

l 

(l+.-)~=e (411) 



44. Corollary. Since 



a 
i 



we have 



4- / V a* 

(1 + ' =((H-»r) =f=H-« + i3+&c. (412) 

45. The number e is the base of Neper's system of 
logarithms, and the logarithms taken in this system 
are called the Neperian logarithms. The Neperian 
logarithms will be generally used in the course of this 
work, and will be denoted by log. as usual. 



178 DIFFERENTIAL CALCULUS. [b. II. CH. II. 

Exponential infinitesimal functions. 

46. Corollary. The log. of (411) is 

4- log. (l + i) = log.e = l (413) 

log. (I + •) = » (414) 

47. Corollary. If in (412) we put for a the value 

a = m i (415) 

and transpose 1 to the first member, (412) becomes 

e mi I __ m {. (416) 

all the terms of the second member except the first being 
omitted, because they are infinitely smaller. 



48. Corollary. If b is taken so that 




b — e m or m = log. 6, 


(417) 


(416) becomes 

frj — I = i log. b. 


(418) 


49. Corollary. If in (416) 




m = 1, 


(419) 


(416) is 

e { — 1 = it 


(420) 



§ 51.] DIFFERENTIALS. 179 

Differences are linear functions. 



CHAPTER III. 



DIFFERENTIALS. 



50. The difference of a function is the difference be- 
tween its two values, which correspond to two different 
values of the variable. 

When the difference between the two values of the 
variable is infinitely small, the difference of the func- 
tion is called its differential. 

The letters a., A.', J.",&,c. t 4. , J. 19 &c, F ., p. 1 , &c, 
placed before a function denote its differences, and correspond- 
ing differences are denoted by the same letters. Thus 

A.fx, a'. fix, &c. 

are corresponding differences of f.(x), fi(x), &c, and these 
differences correspond to the difference 4. x of the variable, so 
that 

*/.* = /. (*+.*. *) — /.*. (421) 

A fix — fi(x+A. x) — fix. &c. 

Differentials are denoted by the letters 3 f 3' f &c, d, d' f &c,. 
D, D', &c. 

51. Theorem. Differences and differentials are linear 
functions. 

Proof. For it is obvious that the increment of the sum Of 
several functions arising from an increase of the variable is 



180 DIFFERENTIAL CALCULUS. [b. II. CH. I1L 

Differences are free relatively to linear functions. 

equal to the sum of the increments of the variable ; or, as it 
may be expressed algebraically, that 

j.if.^f^^Uf. + J.f^^ifJ+j.f.'K-if^fJ) 

= *tf* db J.fi (422) 

52. Theorem. Differences and differentials are free 
relatively to any other linear function. 

Proof. Let f. be a linear function, and (389) gives 

f.(x+J.x) = f. x +f.J.z, (423) 

which, substituted in (421), gives 

J.f.z = f.J.x, (424) 

and this is the theorem to be proved. 

53. Corollary. In equation (399) we may put 

fv = *• 
and it becomes 

(/. + J.)?. = /• + nfn-i j. + &c. (425) 

54. Corollary. If the function f. of (425) is unity, (425) 
becomes 

{l. + j.Y = l. + w^ + U{n ~ A) J. 2 + &c. (426) 

55. In finding differences and differentials, the dif- 
ferences and differentials of the independent variables 
are also independent and may vary or not, as may be 



<§> 59.] DIFFERENTIALS. 181 

Differences of independent variable. 

most convenient. It is usually most simple to suppose 
the differences and differentials of the independent vari- 
ables not to vary, and, adopting this hypothesis, we 
must regard the second differences and differentials of 
the independent variables as constant. 

56. Corollary. By the principle of art. 55, (426) becomes 
when applied to the independent variable x 

(1. + J.) n x=(l.+nj.)x = x + nJ.x (427) 

for we have 

^2 = 0, j. 3 x = 0, &c. (428) 

57. Corollary. Taking the function f. of each member of 
(427), we have 

f.(l. + j.y i x=f.(\. + nj.)x. (429) 

58. Corollary. Equation (423) gives, by transposition and 
omitting x, 

f. + J.f. = f.(l.+ J.) (430) 

or (1.+^)/- =/..(!• +^) (431) 

so that the functions/, and {l. + J.) are relatively free, and 
we have, by (429), 

( 1 - + ^r/^=/.(l. + ^)^=/.(l. + n^.) a ; 

= f.(x-\-?iJ.x) (432) 

59. Corollary. Since the linear function j. may be subject- 
ed to all the forms of algebraic calculation, it may be sub- 
stituted for a in (412), and gives 

16 



182 DIFFERENTIAL CALCULUS. [fi. II. CH. III. 

Differential coefficient. 

eJ.= (l. + j. + ^.l + &,c.y, (433) 

and in like manner 

e"J. = I 1. + n ^•+ T -^- + &c. \ . (434) 

60. The quotient of the differential of a function 
divided by the differential of the variable is called the 
differential coefficient of the function ; the differential 
coefficient of the differential coefficient is the second 
differential coefficient , and so on. 

Differential coefficients are denoted by d c . ; thus 

d,f.x = *g (435) 

d?.f.x=:d c .d c .f.x; (436) 

or, since d % is independent of x, 

e/.*=^ = ^- (437) 

61. Theorem. The differential coefficients of con- 
tinuous functions are finite functions of the variable, 
independent of the differential of the variable. 

Proof. I. Let BC (fig. 47.) be the curve which denotes 
the function /. x, so that if A is the origin, we have, for 

AP = x, PM=f.x, 

and if we take 

PP> — d.x — MR, PP" —d'.x— MR', 



§ 63.] DIFFERENTIALS. 183 

Orders of differentials. 

we have 

d.f.% — PM' — PM = MR, 

d'.f. x = P"M" — PM— M'R', 
MR _ _ M >R< 



MR ' c J MR' 

But since MM' M" is an infinitely small portion of the curve, 
it may be regarded as a straight line, and we have 

MR __ 31" R ' 
~MR ~~ MR* 

or d c .f.x = d' c .f.x; 

that is, the value of the differential coefficient does not change 
with that of the differential. 

II. The differential coefficient is, in general, a finite func- 
tion, for the ratio M' R : MR, which represents this function, 
is the tangent of the angle M' MR, by which the curve is 
inclined to the axis AX. 

62. Corollary, We have, by (435) and (437), 

d.f.xz=d c .f.x.dx,d 2 .f.x=d?.f.x.dx 2 , (438) 

so that if dx is an infinitesimal of the first order, df. x is, by 
art. 30, of the same order, d 2 f. x is of the second order, and 
so on. 

Differentials may then be regarded as infinitesimals 
of the same order with their exponents. 



63. Corollary. If we put 

nt n — 

dx 



ndx — h, so that n = — -, (439) 



184 DIFFERENTIAL CALCULUS. [b. II. CH. III. 

Taylor's and Mac Laurin's theorems. 
and put d. for J. in (432), we have 

(]. + <*.)»/.* = /•(* + A); (440) 

or developing, as in art. 42, and putting 



d, _ d*. 

Tx 

h 2 d 2 



*-=j&*- = 3p (44I > 



(l.+ ftd- + -1^+ &c.)/.z =f.(x+h). (442) 
But, by (434), 

i.+A^.+4-4-+ &c - = e *' <,c - ( 443 ) 

1 • ii 

whence 

e**-/. x — f. (z + h). (444) 

64. Corollary. When, in (442) and (444), we put 
x = 
they become 

(l. + ^ c .+^f- + &c.)/.0 = /.A (445) 

e™ c f.0=zf.h (446) 

and if we now put x for h, we have 

( X +^ C + ^f + &c.)/0=/.x (447) 

e* d <-f.0 = f.x. (448) 

Equation (442) is called Taylor's theorem, of which 
(444) is a neat form of writing, and (447) is called 
Mac Lanrirts theorem. The great use of these the- 
orems will be seen in the sequel. 



§ 68.] DIFFERENTIALS. 185 

Increasing or decreasing function. 

65. Corollary. If we put, in (442) and (444), , 

h ±= j. x, 
they become, by (431), 

(l + j. x d c .+&c.)f.z=f(l. + j.)x=(l.+J.)f.z (449) 
eJ.xd c ,f. x = (1. + J.)f. x. (450) 

66. Theorem. When the differential coefficient of a 
continuous function is finite and positive, the function 
increases with the increase of the variable ; but if the 
differential coefficient is negative, the function decreases 
with the increase of the variable. 

Proof. For the differential coefficient is the ratio between 
the differential of the function and that of the variable, and is 
therefore positive when both these differentials are positive, 
and negative when one is positive and the other negative. 

67. 'Corollary. If the variable increases from any of 
its values, for which the function vanishes, the function 
must be positive if the differential coefficient is positive, 
and negative if the differential coefficient is negative : 
that is, the function has the same sign with the differen- 
tial coefficient. The reverse is the case if the variable 
decreases. 

68. Theorem. The greatest value of the differential 
coefficient of a continuous function, which vanishes with 
the variable, and extends to a given limit, is larger 
than the quotient of the greatest value of the function, 
divided by the corresponding value of the variable; and 

16* 



186 DIFFERENTIAL CALCULUS. [b. II. CH. III. 

Greatest and least values of differential coefficients. 

the smallest value of the differential coefficient is smaller 
than the smallest value of this quotient. 

Proof. Let f. denote the given function which vanishes 
with the variable, let %' be the limit of the variable to which 
the function is extended, and let A and B be respectively the 
greatest and the least of the values of the differential coeffi- 
cient, so that 

A — d c f. x and d c f. x — B 

are positive. But these two quantities are the differential co- 
efficients of the functions 

A x — f. x and f.x — B x, 

both of which vanish when x is zero, and are, therefore, by 
art. 67, of the same sign with their differential coefficients 
when x is increasing and positive, and of the opposite sign 
when x is decreasing and negative. Hence these two func- 
tions are of the same sign with x, and their quotients, divided 
by x, must be positive, thatjs, 

X X 

are positive. It follows, then, that A is greater than the quo- 
tient of/, x divided by x, and that B is less than this quotient. 

The truth of this proposition may be exhibited geometri- 
cally. Thus, if AMB (fig. 48.) is the curve which represents 
f. x, so that, for 

AP =z x, we have MP z=f. x ; 
if the curve at M is produced in the straight line MT, 
we have, by art. 61, 

f/ c ./.x = tang. MTX, 



$ 69.] DIFFERENTIALS. 187 

Greatest and least values of differential c: 

and. by joining AM, 

ting. MAX =^ = f±. (451) 

Now, M 1 AX being the greatest value which the angle 
MAX has in this curve, it is evident that in proceeding from 
Jfj to .1 the curve must be incliued to the axis AX by an 
angle greater than M X AX\ so that at 31. for instance, the 
value of tang. 3ITX. or of d.f. x. is greater than tang. 31 1 AX, 

the greatest value of — : and. therefore. ^4. the greatest 

x ■ ' 

/• 

value of d. f. x. must be greater than any value of "— — . 

x 

Again. M Z AX being the least value of MAX i we see that 
in proceeding from 3I 2 to A the curve must be inclined to the 
axis .4 J17 by an angle less than 3I AX, so that at 31 , for in- 
stance, we have 

tang. M T X < tang. 3I 2 AX : 
whence it follows as above, that B is less than any value of 

-— . 

i 

69. Theorem. If a function / and its differential 
coefficient are continuous, and if the function vanishes 
when the variable is zero, there is. for every value of 
the variable x, a value of 6 less than unity, which satis- 
fies the equation 

f.x = xd : f(~x). (452) 

Proof. If A and B are respectively the greatest and least 
values of d.f. x, contained between the limits 

x = 0, and x =z x, 



188 DIFFERENTIAL CALCULUS. [b. II. CH. III. 

Ratio of functions. 

it follows from the fact, that d c f. x is continuous, that it must 

assume every possible value between A and B } while x varies 

f '. x 
from to x. But the value of — — is included between .4 and 

x 

By and, therefore, there is a value of %' less than x such that 

4/.*' = 4-*; 

or f.x — x d c f. x'. (453) 

But since %' is less than x } if we put 

x> 
6 = — or x' = 6 x, 
x 

we have fl less than unity, and (453) becomes (452). 



70. Corollary. If in (452) divided by x, we suppose x to 
be such a function f x of a new variable Zj as constantly to 
increase with the increase of x x and to vanish with x lt so 
that 

and take & lf so that 

fl-»=/ 4 r(*.i.*i)j 

then a j is evidently less than unity, and if we suppose 

(452) becomes, by dividing by x, 

or 

F.x 1 _ dF. («,«,) _ <*. J- («,*.) . 



<§> 71.] DIFFERENTIALS. 189 

Ratio of functions. 

or, omitting the numbers below the letters, which are no longer 
needed, and put x[ == x 

F. x d c F. (6 x) d c F. x 1 



f.x ~ d e f.(6x) d c f.x< 



(454) 



71. Corollary. If the n first successive differential co- 
efficients of the functions F. and f. were continuous, 
and all but the n th vanished, (454) would, give 

F. x d c F. x' d] F. x" d n c F. x n 



/. x ~ d c f. x' - dif. x" - d: /. x; 

in which x, x', x", &,c. are decreasing, so that if we put 



(455) 







x n 
n X 




we have 










F.x 


d:F.(6„ 


*) 




f>* 


~ </-.(«» 


«>• 


in which 


is less than 


unity. 





(456) 



190 DIFFERENTIAL CALCULUS. [b. II. CH. IV. 

Complete differential sum of partial ones. 



CHAPTER IV. 

COMPOUND AND ALGEBRAIC FUNCTIONS. 

72. Theorem. The differential of a compound func- 
tion of several simple functions, is equal to the sum of 
its partial differentials arising from allowing each 
simple function to vary by itself, independently of the 
other simple functions. 

Proof. Let f.,fJ be simple functions, of which </>. is the 
compound function. We will denote by d f d f , the partial 
differentials, supposing/.,/' respectively to vary by themselves. 
We are to prove that 

d f . (/.,/.') = d f . f. (/.,/.') + d f .. f. (/.,/.') (457) 

Now we have, by definition, 

d f . f. (/,/.') = f. (/. + <*/.,/.') - ,. (/.,/.'), (458) 

or, by transposition, 

f. (/.,//) = f, (/. + df.,/.') - d,. 9 . (/.,/.') 

The differential of this last equation, supposing f.' to vary, is 
4/ • * (/•>/•') = <*/.• v (/• + df.,fJ)-d?,f.< * (/•./•')• (459) 
But, by definition, 
d / ,f.{f.+df.,f.')=f.(f.+df.,f.'+df.') - <p.(f.+df.,f.') (460) 



§ 76.] COMPOUND AND ALGEBRAIC FUNCTIONS. 191 

Differential of product. 
which, substituted in (459), gives 

d t , 9 . (/.,//) = *. (/. + df.jj + dfi) 

- 1- (/• + df.J.') - a.*/.,/.' v. (/.,/.'). (46T ) 
The sum of (458) and (461) is 

df. T (/.,/•') H-4.-; f- (/-/•') = f-(f-+df.,f.'+df.') 

— 9. (/,/•') - df.,,. f . (/.,/.'). (462) 

But we have 

d v . (/.,/.') - r (f. + df.,f.<+df')-<e. (/.,/.<) (463) 

which, substituted in (462), gives (457) by omitting the last 
term, because it is a second differential, and therefore an infi- 
nitesimal of the second order. 

73. Corollary. Equation (457) is written by omitting the 
function, 

d. = dj, + d r . (464) 

74. Problem. To differentiate a f. x. 
Solution. We have, by definition, 

d.{af. x) z= af. (x -j- dx) — af.x 

= a[f.(z+dz)—f.x] = ad.f.z. (465) 

75. Corollary. We have then 

d.(ax) — adz. (466) 

76. Corollary. We have also 

d f .(f.x.fJx) =f.'zdf.z 

d / ,.(f.x.f<x)=f.xdf:z 



192 DIFFERENTIAL CALCULUS. [b. II. CH. IV. 

Differential of power. 

and, by (464), 

d.(f.z.f>z)=:f>zdf.z+f.zdf>z. (467) 

In the same way, if u and v are functions of x, 

d.(uv) = udv -\- v du. (468) 

77. Corollary. Equation (467), divided by f. x.f.'z, is 

f.z.f.'z ~ f.z T f.'x ' { > 

78. Corollary. From (469), it follows that 

d.jf.z.f'zJJ'z^ cLfz_ d^z 
f.x.fJx.f»x... f.z ^ /.'* T V > 

79. Corollary. If in (470) we have the rc functions 

f.x=f.'z—f"z — &c. (471) 

(470) becomes 

*W£»JM£f (472) 

which, freed from fractions, is 

d(f.x)» = nXf.x)*-idf.x. (473) 

Hence, fo differentiate any power of a function, multi- 
ply by the exponent and by the differential of the func- 
tion, and diminish the exponent by unity. 

80. Scholium. The proof of (472) and (473), which is 
given in art. 79, is limited to positive integral exponents, but 
may be easily extended. 



§ 80.] COMPOUND AND ALGEBRAIC FUNCTIONS. 193 

Differential of power. 

I. Proof of (472) and (473) for fractional exponents. Let 

n be a fraction 

m 

n = — 

m' 

and let cp.x = (fx) n 

so that (<p.x) m ' = (/. x) m . 

Equation (472) gives 

m! d ((p. x) m d (f x) 





(p. x 


" /.* 


or 








<P.x ' (fx) n 


m d{f x ) n d(fx) 
" «*'" f *> f* 



which includes (472), and consequently (473). 

II. Proof of (472) and (473) for negative exponents. Let 
n be negative 

n = — m 

1 



and let (p . z =z (f x) 



(f.*r 

so that (p. x (fx) m = 1. 

The differential of which gives, by (470) and (472), 

dcp.x d(fx) m dcp.x mdf.x 

tp.x* (f x) m v. x f. x 

whence 

d(fx) n _ m dfx_ n df x 

Jfxf- ~fx~-~fx~ 

which is the same as (472), and therefore includes (473). 



17 



194 DIFFERENTIAL CALCULUS. [b. II. CH. IV. 



Binomial theorem. 



81. Corollary. Equation (483) gives 

d.x m =mx m - l dx (474) 

and d c . % m = mx m - 1 (475) 

d 2 c .x m — md c . x m - 1 = m(m— ])z m - 2 (476) 

d\. x m —m(m — l)(m — 2) x™~* , &c. (477) 

If now we substitute x m for /. x in (442), we have 

(x + hr = x™ + hd c .x>»+^-d*.x™ + &c. 

— X ™_L_ mx™-Vi+ m ^™~^ x™- < *W+&c,. (478) 
Avhich is the binomial theorem. 

82. Examples for Differentiation. 

1. Differentiate ax m -\-b. 

Ans. The differential coefficient is m a x m ~ l . 

d f.x 

2. Differentiate \/(f. x). Ans 



3. 


Differentiate — -. 


4. 




~ (/•*)"• 


5. 


f. X 

Differentiate Vt-. 


6. 


Differentiate ) J ' ' 



Ans. — 



2 \Z(j. *)" 
df.x 



(f.*) 2 
nadj 
[f^f 
f.'x.df.x—f.x.df.'x 



n a df. x 

Ans - ~ OG^+i 



Ans. 



(fjxy 



(f.x) m - 1 (m f.'x.df.x — m'f.xdf.'x) 



<$> S2.] COMPOUND AND ALGEBRAIC FUNCTIONS. 195 

Differential of polynomial. 

7. Find the successive differential coefficients of 

a + b x + c x 2 + + 31 1". 

Ans. The first is b -f 2 c z . . . + m Mx™- 1 ; 

the second is 2c , . . -f »i (»i — lJJfi" 1 " 2 ; 

the ??* th is ro(w— l)(m — 2)... 1.3/; 

the (w+l) st is 0. 



196 DIFFERENTIAL CALCULUS. [b. II. CH. V. 

Differential of exponential. 



CHAPTER Y. 



LOGARITHMIC FUNCTIONS. 



83. Problem. To differentiate a*. 
Solution. We have, by definition, 

d a* =z a x + dx — a x = a x a dx — a* 

— a x {a dx —l). - (479) 

But, by (418), 

a d * — l) = log. a. dx; (480) 

whence 

da* — log. a. a x dx. (481) 

84. Corollary. Hence 

d c a x = log. a. a x 

d*. a x = log. a. d c a x — (log. a) 2 a x 
d n c .a x — (\og.aya x (482) 

and by making x = 

a° = 1 
d c . a — log. a 
d%. «o — (log. «) 2 
dj. a° — (log. a)". (483) 






§ 86.] LOGARITHMIC FUNCTIONS. 197 

Differential of logarithm. 

If now we put in (447) 

£ x = a? 
we have 

a' = 1 + log. a. x + (1 ° g ; *• * )2 + &c. (484) 

1 . — ■ 

85. Corollary. If we have a — e, we have 

log. e == "1, 
e* = d c . e x — d] . e x z= d n c . e z (485) 

1 = d e . e° =s «£ e° = £. e° (486) 

e* = 1 + x + ^ + &c. (487) 

and (487) is the same with (412). 

86. Problem. To differentiate log. x. 
Solution. "We have, bv definition, 

d. log. x — log. (x -\- d x) — log. x 

= 'og.^- = log.(l + ^). (488) 

But, by (414), 

and, therefore, 



d. log. x = — . (489) 



17* 



198 DIFFERENTIAL CALCULUS. [b. II. CH. V. 

Development of logarithm. 

87. Corollary. Hence 

d c .\og. x — - (490) 

dl log.x = d c .-= — - 

a* log. £ = — 

the upper sign being used when n is odd, and the lower when 
n is even. 

If now in (442) we make 

/• = log- 
we have, 

log.(* + A) = log.*+i- — + 3^-&c. (492) 

88. Corollary. If in (492) we put 

X sac I, 

we have 

log. (1 +A)= i — J/i 2 + i^ 3 — 'f* 4 4" &c - ( 493 ) 

89. Examples. 

1. Differentiate log. [x +\/(l +x 2 )]. 

dx 
Ans 



2. Differentiate (log. x) n . Ans 



V(l + * 2 )' 

n (log. x) n — 1 dx 



§ 89.] LOGARITHMIC FUNCTIONS. 199 

Differential of 2d and 3d logarithms. 

dx 
3. Differentiate loo;. 2 x. Ans. 



4. Differentiate locr. 3 x. Ans. 



x log. x' 
dx 



xlog.x. log. 2 x 

5. Differentiate a^ x . 
Solution. Let y = b* 

and we have a^ x = a y 

whence d. ab x — d. a y = log. a. a y '. d y 

But dy = d.b x = log. 6. 6*. ds 

so that d.ab*= log. a. log. 6. <$ x . b x d x. 

6. Differentiate x y . 
Solution. Equation (464) gives 

d.x y — d x .x y -\-d y .x y 
But by (473) and (481) 

d x .x y = yxv — 1 dx 
d y .x y =z log. x. x y dy 
so that 

d.x y = y xy—* 1 dx -\- log. x. x y dy. 



200 DIFFERENTIAL CALCULUS. [b. II. CH. VI. 

Differential of sine and cosine. 



CHAPTER VI. 



CIRCULAR FUNCTIONS. 



90. Problem. To differentiate sin. x. 
Solution. We have, by definition, 

d. sin. x z=z sin. (x -\- dx) — sin. x. 
But, by trigonometry, 

sin. (a; + dx) = sin. x cos. d x -f- cos. x sin. d x, 
cos. d x =. 1 , sin. d x = d x, 
so that d. sin a: = cos. Z. ds. (494) 

91. Problem. To differentiate cos. x. 

Solution. Substitute in (494), %n — X) and it becomes 
d. sin. ( J 7T — &) = cos. ( J n — x) . d ( J tt — «). (495) 
But we have 

sin. ( J 7T — x) = cos. x, cos. ( J ^ — x) = sin. x 
d (i™ — x) = — e? z, 
which, substituted in (495), give 

c?. cos. x = — sin. x dx. (496) 



§ 93.] CIRCULAR FUNCTIONS. 201 

Development of sine and cosine. 

92. Corollary. Equations (494) and (496) give 
d c . sin. x = cos. x 

d*. sin. x = d c . cos. af— — sin. x 

d 3 c . sin, x — d z c . cos. x — — d c sin. x = — cos. x 

c/ 4 c . sin. x =z d 3 c . cos. z = — d c cos. a; == sin. a; 

d n c . sin. 2; = dj -1 . cos. x= d n ^. sin. a, (497) 

so that the four values of all the successive differential co- 
efficients of sin. x and cos. x are alternately cos. x, — sin. x, 
— cos. x, and sin. x. 

Hence, making x ~ 0, 

we have, when n is even, 

d n c . sin. = d n ~ l cos. — 0, (498) 

but when n is odd 

d n c . sin.O = d n c ~ l cos. = ± 1 ; (499) 

the upper sign being used when n — 1 is divisible by 4, and the 
lower sign when n -\- 1 is divisible by 4. 

These values, substituted in (447), give 

siD - x = x ~ m + rdfcus + 1^X6^+ &c - ( 50 °) 

co " =1 -i¥ + iin-uu5.6+ fa '' 5M » 

93. Problem. To differentiate tang. x. 

Solution. We have, by trigonometry and example 6, of 
art. 81, 

sin. x cos.xd c s\n.x — sin. xd c cos.x 



d c . tang. x=z d c 



cos. 2 x 



202 DIFFERENTIAL CALCULUS. [b. II. CH. VI. 

Development of tangent. 

whence, by (494), (496), and trigonometry, 

_. cos. 2 x -4- sin. 2 x 1 >»«1*« 

d c \ tang, x = X = — . (502 

COS. 2 X COS. 2 X ' 



94. Corollary. Equations (502) and (496) give 

2d c cos. x 2 sin. x 2 tang, x 



d 2 c . tang, x 



3 



COS. * X COS. ° 2 COS.* X 

or cP c . tang, x = 2 tang. x. </ c tang. £ 

dj tang, x — 2 tang, a; d\ tang, a; -f- 2 (d c tang, x) 2 

c/* tang, z == 2 tang, a; ^ tang, x -}- 6 tf c tang, x d* tang, x 

^ tang, x z= 2 tang, x rf* tang, x -(- 8 c? c tang, x ^ tang, x 

+ 6(rf c 2 tang. x) 2 
rf"tang. xz= (r/ c . tang, x -f- d c . tang. x)™- 1 . (503) 

in which the exponents are to be annexed to the d c . ; when 
the exponent is zero, as in the first and last terms, the d c . is 
omitted, and tang, x retained ; and as the terms equally dis- 
tant from the two extremities of the developed series are alike, 
they may be added together. 

By making x = 0, 

these equations give 

d c tang, x = 1 

d\ tang, x z= 

d\ tang, x = 2 

d]. tang, x — 

d 5 c tang, x — 16. 



§ 97.] CIRCULAR FUNCTIONS. 203 



Differential of negative sine and tangent. 



Hence, by (447), 



2 16 

= ^ + i^ 3 +T 2 ^ 5 + &c. (504) 



95. To differentiate arc sin. 2: =1 sin.C -1 ^. 
Solution. Let y = sin.[— ^x 

or sin. y = x 

so that by differentiation 

cos. y.dy = d x = d y \/(l — sin. 2 y) =dy\/(l — x 2 ) 
and 

4. sinJ-n * = J, y = g = ^__ = (1-%-J. (505) 

96. Corollary. By the same process we should find 

d e . cos.C- 1 ] x = d c arc cos. z — — (1— x*)-h. (506) 

97. Problem. To differentiate tang.C— 1 ^ = arc tang. x. 
Solution. Let y = tang.C— ^ x 

or tang, y =1 x, 

so that, by differentiation, 

J^ = dx = dysec*y=dy(l+tzngZy)=:dy(l+z*) 

and 

rf c tang.[-i]* = rf e y=f^. = (i+ x 2)-i. (507) 



204 DIFFERENTIAL CALCULUS. [b. II. CH. VI. 

Development of negative sine. 

98. Scholium. By applying Taylor's theorem, the 
values of sinJ— *] x, cos. I— *] x, and tangJ— ^ x, might 
easily be found expressed in series ; but they may more 
easily be found by the following process. 

99. Problem. To develop sin.t -1 ^ in a series arranged 
according to powers of x. 

Solution. Suppose the series to be 

sin.t- 1 ! x — C + C % x + C 2 x* + C 3 x* + C 4 z* + &c. (508) 

in which the number below the coefficient denotes the power 
of x, which it multiplies. We find, then, by differentiation, 
(505), and the binomial theorem, 

(1— x2)-i = C 1 +2C 2 x+3C 3 x2 + 4C 4 i3+& c . 

_ . 1.3 . , 1.3.5 

X 6 -\-&LC. 





-*-r»* i a.r ' 2.4. 


whence 






c, = i 




<V-;*-i™* 




C5 — T -274- ¥?J 


and, 


C„ =0 


when n is even, 


and if in (508) we put 




x=z 


it becomes 






c = o. 



Hence, by substitution, 

sin j^ x = x + ix3+A»* + &c. (509) 



§ 102.] CIRCULAR FUNCTIONS. 205 

Development of negative eosine and tangent. 

100. Corollary. In the same way, if we put 

cos.t- 1 ] x = C + C x x + C 2 %z + &c. (510) 

we find all the coefficients except C Q to be the negative of 
those for sin.t - J ] x, and if in (510) we put 

x — 
we find 

C =. arc. cos. = % n t 
whence 

cos.t-^x^r \™— x — £x 3 — &X* — &C (511) 

101. Corollary. If in (511) we put 

x— 1 
we have 

COS.C- 1 ! 1 — = 1 7v — 1 — £ — & — &c. (512) 

and if (512) is subtracted from (511), term from term, it gives 
cos.[- 1 Jx = (l — x)+^.(l— x3)-f ¥ 3 7 (i_ X 5)^_ &c . (513) 

102. Problem. To develop tang.t—^a? in series, ac- 
cording to powers of x. 

Solution. Suppose the series to be 

tangJ- 1 ] x = C + C 1 z + C 2 x2 + &c. (514) 

we find, by differentiation, (507), and the binomial theorem, 
(1 + z2)-i = C 1 +2C 2 z + 3C 3 x2 + 4C' 4 x3+&c. 
— 1 _ X 2 _|_ X 4 __ x e _j_ &, c . 
whence C 2 = 1 

<Y=.-* 
c. = i 

and when w is odd C n =. i -, 

18 



206 DIFFERENTIAL CALCULUS. [b. II. CH. VI. 

Differentials of circular functions. 

the upper sign being used when n -\- 1 is divisible by 4, and 
the lower sign when n — 1 is divisible by 4 ; but when n is 
even or zero, 

c„ = o. 

Hence, by substitution, 

tang.C- 1 ^ = x — £z 3 _|_.i x s __ i X 7 _|_ &, c . (515) 



103. Examples. 

1. Differentiate sec. a;. Ans. tang, x . sec. x d x. 

2. Differentiate sin." 2;. Ans. nsm. n — l x cos.xdx. 

3. Differentiate log. cos. z. Ans. — tang. xdx. 

, t^'/y. • 1 / /I +sin.z\ 

4. Differentiate log. \/ I : I. Ans. sec. xdx 

6 ^ \1 —sin. xf 

~ T^m 1 r n / 6 + flCOS. X \ 

5. Differentiate -~^ — — r cos.t- 1 ! ( — \— 1. 

V(a 2 — b 2 ) \ a -\- cos. x J 

a dx 

Ans. 



a -J- b cos. x 
dx 



6. Differentiate cot. x. Ans. 

sin. 2 2; 



§ 104.] INDETERMINATE FORMS. 20f 

Value of fraction, when both terms vanish. 



CHAPTER VII. 



INDETERMINATE FORMS. 



104. Problem. To find the value of a fraction when 
its numerator and denominator both vanish for a given 
value of the variable, of which they are both continu- 
ous functions. 

Solution. Let the fraction be 

§= < 516 > 

and let x be the value of x, for which the terms vanish ; let 
h =z x — x Qi or x — x -f- h, (517) 

the given fraction becomes then 

J - ('.+»> (518) 

which is a fraction, both whose terms vanish for 

h = ; (519) 

so that in (456), h being the variable instead of x f we have, 
if all the differential coefficients of the terms vanish, up to 
the ftth ? w hen 

h = 0, 

f.(x + h)-d" e .f.{* +6 n hy { ™ } 



208 DIFFERENTIAL CALCULUS. [b. II. CH. VII. 

Fraction of which both terms vanish. 
which when h = 

becomes :&&- = ^lElll (521 * 

so that the value of the given fraction is obtained by 
differentiating its numerator and denominator, until a 
differential coefficient is obtained, which does not van- 
ish for the given value of the variable. 



105. Examples. 

1. Find the value of — : . when x = 0. 

sin. x 

Solution. We have 

c o_ e -o d c .e°—d c .e-° e°+c-° 1 + 1 



= 2. 



sin. d c . sin, cos. 1 

sin ( x^\ 
2. Find the value of — — — -, when x = 0. Arts. 0. 



sin. x 
3. Find the value of — - — , when x = 0. Ans. oo. 



4. Find the value of — — -, when x = 1. Ans. 1. 

x — 1 



5. Find the value of -, when x = a. Ans. na 71 — 1 . 



1 cos X 

6. Find the value of — - — , when x = 0. Ans. J. 



x — - sin x 
7 Find the value of — , when x=z 0. Ans. £. 



§ 106.] INDETERMINATE FORMS. 209 

Fraction of which both terms are infinite. 

e x e — x 2 X 

8. Find the value of : , when x=z0. 

x — sin. x 

Ans. 2. 

9. Find the value of — , when x = 0. 



















Ans. 


h 


10. 


Find the value of 


sin. 

sin 


.9 ' 


when 


<P 


= 0. 




Ans. 


m. 


11. 


Find the value of 


sin. 
sin 


m (p 

> 


when 


g> 


= 7Z, 


and 


m is an 


intes 


rer. 



















Ans. — m cos. m n, so that when m is even, the answer is 
— m ; and when m is odd, the answer is m. 

106. Problem. To find the value of a fraction, when 
both its terms become infinite for a given value of the 
variable. 



Solution. 


Let the numerator of the fraction be 


Y and the 


denominator 


y, the fraction 


is 






Y _ 

y ~ 


y- 1 

Y-\ 


(522) 



But when y and Y are infinite, their reciprocals, y~ l and 
F —1 , vanish, and we have by the preceding art. 

Y _ d,y~i -y-*d c y _ Y*d c y 
y " d c . F-i — F-» di Y ~" y* d> c Y 

Y 2 

and, dividing by — - 

y-i:y or J = dTy> (523) 

18* 



210 DIFFERENTIAL CALCULUS. [B. II. CH. VII. 



so that the value of the fraction may be found by dif- 
ferentiating both its terms ; and if both the terms of 
the fraction thus obtained are infinite or zero, the dif- 
ferentiation may be continued, until a fraction is obtain- 
ed, of which both terms are not infinite or zero ; and 
equation (521) applies to the present case as well as to 
the preceding one. 

107. Examples. 



1. Find the value of °* , when i = 0. 
cot. x 






Solution. We have 

log. d c . log. sin. 2 

cot. "~ gL cot. 0- — 



2 sin. cos. = 0. 



„ _, , , _ _ cot. m cp ] 

2. f ina the value ol , when <p = 0. Ans. — 

cot. ^ m 



108. Problem. To find the value of a product of 
two factors, when one of the factors become infinite 
and the other factor becomes zero for a given value of 
the variable. 

Solution. Let y and Y be the two factors, and we have for 
the given product 

V Y = -^T (524) 

so that it is equal to a fraction, of which both the terms 



§ 110.] INDETERMINATE FORMS. 211 

Infinite or zero powers. 

are infinite, or both are zero, and the value of this frac- 
tion may be found by the preceding articles. 

109. Examples. 

1. Find the value of x a e~ x , when x —z oo. Ans. 0. 

2. Find the value of x (log. x) n , when x z= 0. Ans. 0. 

3. Find the value of z a log. x, when x = 0. ^ins. 0. 

4. Find the value of x cotan. a, when x — 0. -4ws. 1. 

110. Problem. To find the value of a power, when 
the exponent and the root are both such functions of a 
variable, that they assume, for a given value of the 
variable, one of the forms 0°, oo°, or l 30 . 

Solution. Let the power be 

z 3= F* (525) 

and we have, by logarithms, 

log. z = y log. F, (526) 

so that in either of the given cases log. z is equal to a 
product, of which one of the factors is zero, while the 
other is infinite ; its value may therefore be found by 
the preceding articles ; and when its value is found, we 
have 

z = e l °s^ (527) 

or Yy = ey l °s-r (528) 



212 DIFFERENTIAL CALCULUS. [b. II. CH. VII. 

Infinite or zero powers. 

111. Examples. 
1. Find the value of x x , when x =z 0. 
Solution. Since log. = 
we have 0° = e° = 1. 

% Find the value of (e x — 1)*, when % = 0. Ans. 1. 

3. Find the value of cot.^ q>, when gj = 0. Ans. 1. 

4. Find the value of cos. <jD cot -g>, when cp = 0. -4ws. 1. 



§ 113.] MAXIMA AND MINIMA. 213 

Maxima and minima of function of one variable. 



CHAPTER VIII. 



MAXIMA AND MINIMA. 



112. A value of a function, which is greater than 
those immediately preceding and following it, is called 
a maximum of the function ; while a value, which is 
less than those adjacent to it, is called a minimum. 

113. Problem. Find the maxima and minima of a 
continuous function of one variable. 

Solution. Let f. be the given function, of which all the 
differential coefficients, inferior to the nth, vanish, for a given 
value x of the variable. Then 

/•(*«+*)-/.*. (529) 

is a function of h, which vanishes with h ; and its differential 
coefficients, inferior to the rath, also vanish with h, for they 
are, when h is zero, 

«• /• (*. + *) = « /• *o = 0. (530) 

Moreover h n is a function of h, all the differential coeffi- 
cients of which, inferior to the rath, vanish with h, and the 
nth differential coefficient is 

d n c .h n — 1.2.3 n. (531) 

If now, in (456), these two functions are substituted, and if 
h is regarded as the variable, (456) becomes 

I 



214 DIFFERENTIAL CALCULUS. [b. II. CH. VIII. 

Rule for maxima and minima. 

¥ - 1.2.3....» {5i2 > 

But if h is taken infinitely small, x Q -\- & n h differs infinitely 
little from x , and the & n h may be neglected, so that (532), 
multiplied by h n , becomes, by transposition, 

/• (». + *) =/• *o + u 8> &" . . - *.•/• V (533) 

By changing /* into — A, (533) becomes 

/• (*. -*) =/■ *o + 1, a"^. r . , <•/• '.■ ( 534 ) 

If, then, w is odd, of the values of the given function, 
which are adjacent to f.x Qi one is greater, while the other is 
less than/. x , so that this value does not correspond to a 
maximum or minimum. 

But if n is even, the values which are adjacent to f. x are 
both greater than fx , when the n ih differential coefficient is 
positive; and they are both less than f. x Q , when this coeffi- 
cient is negative. 

Hence, to obtain the maxima and minima of a given 
function, find the values of the variable which reduce the 
first differential coefficient to zero. Each of these values 
of the variable must be substituted in the succeeding 
differential coefficients, until one is arrived at, which 
does not vanish. 

Jf the first differential coefficient which does not van- 
ish is even and positive, the corresponding value of the 
function is a minimum; but if the coefficient is even 
and negative, the value of the function is a maximum. 



<§> 115.] MAXIMA AND MINIMA. 215 

Case when differential coefficient is infinite. 

114. Scholium. When the function, varying with the in- 
crease of its variable, passes through a maximum or minimum, 
its difference must change its sign. If this difference is con- 
tinuous, it can only change its sign by passing through zero ; 
but if it is discontinuous, it may also change its sign by pass- 
ing through infinity, and the first differential coefficient which 
does not vanish must in this case be infinite. 

The case, therefore, in lohich the first differential co- 
efficient, which does not vanish, is infinite, deserves 
particular examination; and all other cases of discon- 
tinuity are to be considered by themselves, but they are 
rarely of much interest. 



115. Examples. 

1. Find the maxima and minima of the function x 3 -f- a x -f- b. 
Solution. The differential coefficients are 

d c .(x* + ax-\-b) = 3z2 _|_ a 

d%(z* ^j- ax + b) = 6x 
d\. (x*3 + a x + b) — 6. 
The first coefficient is zero when 

so that there is neither maximum nor minimum, unless a is 
zero or negative. For this value of x, the second coefficient 
becomes 

db 6 a/— % a, 
which, when a is negative, is positive for the positive value of 



"216 DIFFERENTIAL CALCULUS. [b. II. CH. VIII. 

Examples of maxima and minima. 

x, and negative for the negative value of x. The correspond- 
ing values of the function are 

But when a is zero, the second differential coefficient also 
vanishes, and the third does not, so that there is neither maxi- 
mum nor minimum. 

2. Find the maxima and minima of the function 

a x 2 -\- b x -\- c. 

4 a c b 2 

Ans. The value is a maximum when a is 

4 a 

negative, and a minimum when a is positive ; the correspond- 
ing value of x is 

3. Find a maximum or minimum of the function 

e x -\- e~ x -\- 2 cos. x. 

Ans. The value 4 is a minimum, which corresponds to 
the value 

x — 0. 

4. Find a maximum or minimum of the function 

Ans. The value 2 is a minimum, which corresponds to the 

value 

x — 0. 

5. Divide a number into two such parts that their product 
may be a maximum or minimum. 

Anst The product is a maximum, when the two parts are 
equal. 



$ 115.] MAXIMA AND MINIMA. 217 

Examples of maxima and minima. 

6. Divide a number into two such parts that the product of 
the m th power of the one by the 71 th power of the other may 
be a maximum or a minimum, m and n being positive. 

Ans. The product is a maximum, when the parts are in 
the same ratio as the exponents of their powers. 

7. Inscribe in the triangle ABC (fig. 8.), the greatest or 
the least possible rectangle DEFG. 

Solution. Using the notation of art. 41, example 2, we have 

the surface DEFG = — 7 '-. 

A 

which is a maximum, when 

% =z £ A. 

8. Through a given point C (fig. 15.) draw a line BCD, so 
that the surface of the triangle ABD, intercepted between 
the lines AB and AD, may be a maximum or a minimum. 

Solution. This surface is, by art. 42, example 7, 

% x y sin. A = J (a y -f- b x) sin. A. 

the first and second differential coefficients of which are 

= i (y + x dcV) sin. A = J (a d c y + b) sin. A. 

{d c y + izd 2 y) sin. A = % a d 2 c y sin. A. 



Hence 



j & — y 

d <y= ; n 

x — a 

= bx — ay, 

y = 2b, x = 2a, %y = ^r, 



and the surface is a minimum. 
19 



218 DIFFERENTIAL CALCULUS. [b. II. CH. VIII. 

Maxima and minima of product. 

9. Find on the circle (fig. 33.) referred to the centre B, as 
the origin of rectangular coordinates, the point M, for which 
the product of the coordinates is a maximum or a minimum. 

Ans. The product is a maximum when y — : x. 

10. Find on the ellipse (fig. 35.), referred to its centre as 
the origin, its axes as the axes of coordinates, the point M, 
for which the product of the coordinates is a maximum or a 
minimum. 

Ans. The product is a maximum, when the coordinates are 
in the ratio of the axes. 

116. When the function, of which the maxima and minima 
are to be found, is a product or quotient, the solution is often 
simplified by finding the maxima and minima of its logarithm, 
which evidently correspond to those of the function. 

Examples. 

1. Find the maxima and minima of the function x~ a e bx , 
when a is positive. 

Solution. We have 

log. (x~ a e bx ) — — a log. x -\- bz 

d c . log. (x— a e bx ) = - + & 

dl.\og.(x-«e b *)= £ 

so that the value 

a 

% = T 



$ 116.] MAXIMA AND MINIMA. 219 

Maxima and minima of product. 

corresponds to a maximum. The corresponding value of the 
given function is 



(t)- 



2. Find the maxima and minima of the function 
(z — a) (x — 6)i-2. 

Ans. There is a minimum, when 

_ 2ab 
X ~a+b' 



220 DIFFERENTIAL CALCULUS. [b. II. CH. IX. 

Orders of contact. 



CHAPTER IX. 



CONTACT. 



117. When two curves meet or cut at a given point, 
and, at any infinitesimal distance of the first order from 
this point, are at a distance apart, which is an infini- 
tesimal of the n -f- 1 st order, they are said to be in 
contact at this point, and their contact is of the n th or- 
der. 

When one of the curves is of the first degree or the 
straight line, it is the ordinary tangent. 

When n is zero, there is no contact, but only an in- 
tersection. 

118. Theorem. When two curves are in contact, the 
portion which is intersected between them upon a line, 
drawn at an infinitesimal distance from the point of 
contact, and inclined by a finite angle to the directions 
of the curves, is of the same order of infinitesimals with 
the distances of the curves apart. 

Proof. Let M< 3I Q M and M[ M M 1 (fig. 49.) be the curves, 
A the point of contact, MN their distance apart, and MM X 
the intercepted portion of the line PM. The line M X N may 
be regarded as a straight line, and the angle 31N3I 1 as a 
right angle ; so that, we have 



§ 120.] CONTACT. 221 

Conditions of contact. 

31M, = MN X sec. NM3f 1 . 

But sec. NMM X is finite, as long as the angle MM X N, by 
which MNM X differs from a right angle, is finite ; and, 
therefore, by art. 31, MM X is of the same order with MN. 

119. Problem. Find the algebraic conditions which 
denote that two giveu curves have a contact of the n lii 
order. 

Solution. Let the equations of the two given curves MM M, 
M' 1 M Q M 1 (fig. 49.), referred to rectangular coordinates, be, 
respectively, 

y = f.x (535) 

y =/,.*• (536) 

Then if A is the origin and M the point of contact, we have 
at this point 

AP z=x 0) P M z=zy Q 

or f. x =f x . x or /. x — f ± . x Q = 0, (537) 

and if PP = h 

we have MM, =f. (x Q + h) —f v ( % + h). (538) 

If, now, f. x — f x . x is such a function of h that all its 
differential coefficients inferior to the m th are zero, we have, 
by (533) and (537), substituting /. — f x . for /. 

MM i = T-o^—Z *?■ (/• -o ~fi- *o) (539) 



1.2.3. 



and if we put 



x __ ff-(/-s -/,.s ) _ <%.f.x -d".f x .x 
1.2.3. ...m ~~ 1.2.3. ...m 

19* 



(540) 



222 DIFFERENTIAL CALCULUS. [b. II. CH. IX. 

Condition of curves crossing at point of contact. 

we have 

M31 1 = JC .h m } (541) 

so that if h is of the first order, MM 1 is of the m th order. 
But MM X is to be of the (n-\- l) st order, and therefore we 
must have 

m = n + ], MM X — X .h n + 1 (542) 

X - d < n+1 f' x o- d < n+1 'f i- x o 543^ 

X ° - ~ 1.2.3 (»+lj~~ } 

and all the differential coefficients of f. x — f x . x, inferior to 
the (w-f- l)st ? must be zero; that is, we must have, when m is 
less than n -\- 1, 

d-f.x -d:f 1 .x = (544) 

or d?/. x =z d m c f x . x . (545) 

120. In the same way, it may be proved that if the 
equations of the two curves are expressed in the polar 
coordinates of art. 45 ; so that they are 

r — F. <p, r = F x . q> (546) 

we must have, when m is less than n + 1. for the point 
r Q , <p of contact, 

«f- F. % z=z d? F v To- (547) 

121. Theorem. Two curves, which are in contact, 
cross each other at the point of contact, when the con- 
tact is of an even order ; but, if the order of the contact 
is of an uneven order, they do not cross. 

Proof. For, when h is negative in (541), the sign of MM X 
is the same as when h is positive if n is uneven, but is the 



§ 123.] CONTACT. 223 

Tangent. 

reverse if n is even ; that is, if n is even, the point M' (fig. 49.) 
is nearer AP than M x \ while the point M of the same curve 
with M 1 is farther from AP than M x of the same curve 
with M x ' ; and if n is uneven, the reverse is the case, as in 
(fig. 50.) 

122. Problem. Through a given point upon a given 
curve, to draw a tangent to the curve. 

Solution. Let x and y be the coordinates of the given 
point, and (535) the equation of the given curve. If r is the 
angle which the tangent makes with the axis of x, its equation, 
since it passes through the given point, is by (162), 

y — y — tan s- r ( x — x 'o)- ( 548 ) 

Hence, by differentiation, 

ld c . y — tang, r — d c . y Q = d c .f. x Q (549) 

and the equation of the tangent is 

y — yo — d c-f- * ( x — ^o)- ( 55 °) 

123. The projection PT (fig. 51) of the tangent 
MT upon the axis of x is called the subtangent, and 
if we put 

AT — x 
the coordinates of T are 

y = and x — x f , 



so that 



pr= Io -^4- = A-=i|»-. (55i) 

d c .f.x d c .y d e .f.x 



224 DIFFERENTIAL CALCULUS. [b. II. CH. IX. 

Tangent. 

124. Corollary. When the equation of the curve is express- 
ed in polar coordinates, as in (546), the tangent can be found 
by means of (117). Thus we have 

^ = Jtt — a } or a z=z ^ n r (552) 

</> + « = J ^ — r -\-(p ~^tv— (t — (f) 

r cos. (<p -j-«) z= rsin. (t — <p) z=p f (553) 

and by logarithms 

log. r -f- log. sin. (t — y) =z log. p. (554) 

The differential of which is, by transposition, 

d c r cos. (t — </>) 1 

-^- == - — { = ; rz=cotan. ( T — if). (555) 

r sin. ^ — (/)) tang. (^ — y) 

But in (fig. 51.), we have 

ip = MAP, t = MTP, (556) 

and if we put 

s — AMT =t — <p (557) 

s is the angle which the curve makes with the radius vector 
at the point 31, so that 

d r 
cotan. « == -^— -= d c log. r = d c log. JP. y (558) 

r 

and the equation of the tangent at the point r y is 

r sin. (t — y) — r sin. *. (559) 

125. The perpendicular MI to the tangent at the 
point of contact is called the normal to the curve. 






§ 129.] CONTACT. 225 

Normal. 

If v is the angle which the normal makes with the axis of 
z, we have then, by (549), 

? — i * + r (560) 

tang, v = — cotan. t — = , (561) 

tang.r d c y ' \ ' 

and the equation of the normal is 

y — Vo =tang.*(z — i ) — — -—-(z— x- ). (562) 

126. The projection PI of the normal Mi" upon the 
axis of x is called the subnormal y so that its value, 
found like that of the subtangent, is 

AI— x Q = y Q d c y ==/. z ,d c f.x Q . (563) 



127. Corollary. The lengths of the tangent and normal, 
found from the right triangles MTP and MP I, are 

OS ' 

T= i»/T= V^i 52 + P T 2 ) — ■/*- V[i + «y ) 2 ] (564) 

iV=i»U=: % /(J7^ + P/ 2 )=y oN /[l + Ky ^] 

= y tf c y . (565) 

128. The tangent is called an asymptote, when the 
point of contact is at an infinite distance from the 
origin. 

129. Scholium. It must not be overlooked that in finding 
d c y, x has been regarded as the independent variable. But if 
it were not so, and if some other variable, as u, were the inde- 
pendent variable; then, denoting by d c3 .. y d cu . } the differential 



226 DIFFERENTIAL CALCULUS. [b. II. CH. IX. 

Tangent. 

coefficients taken on the supposition that x, u are respectively 
the independent variables, we have, at once, 



130. If all the terms of the equation of the given curve 
were transposed to its first member, and if Fwere this first 
member, the equation would be 

V=0, (567) 

whence by differentiation, putting V for the value of Fat 
the point of contact 

d c . x V dx + d c . y V dy=0, (568) 

d i/ 
from which the value of -— , being found and substituted in 
a x 

(548), gives by reduction for the equation of the tangent 

^o^.^-^ + ^oFob-rt^O, (569) 

so that the equation of the tangent may be found by differen- 
tiating the equation of the curve and substituting x — x and 
y — y ! for d x and dy respectively. 



131. Examples. 

1. Find the tangent, normal, &c. of the circle. 

Solution. The differential of equation (58) of the circle 
gives for any point x 0f y 01 

z Q +yod c y = 0, 

tang. r = -, tang. v s= -^. 

Vo x o 



§ 131.] CONTACT. 227 

Tangent and normal to circle and ellipse. 
The equation of the tangent, reduced by (5S), is 

y y + x o x = fo + A — R 2 , 

and that of the normal is 

2-0 y — y z = o, 

so that the normal, by art. 109 of Book I., passes through the 
centre, and is the radius. We have also 

yl R 2 — x\ 

the subtangent = — — = 

x o x o 

the subnormal — — x Q 

R y 

the tangent = — — , the normal —z R. 

x o 
Again, from equation (56) we find 

cotan. s ±= d c log. R = 
« = 90° ; 

that is, the radius vector drawn from the centre is perpendicu- 
lar to the tangent. 

2. Find the tangent, normal, &c. of the ellipse, of which 
A and B are the semiaxes. 

B 2 x. A 2 y n 

Ans. tang. x = - — i, tang, r = g^. 

The equation of the tangent is 

A 2 1/o ij + B 2 x x = A 2 B 2 ; 

that of the normal is 

-B 2 x y + A 2 y x=(A 2 -B 2 )x o!/o . 

A 2 y 2 A*—x 2 A 2 
The subtangent ±= ^r 9 - — - = &o • 

Jt> Xq X q Xq 



228 DIFFERENTIAL CALCULUS. [b. II. CH. IX. 

Tangent and normal to hyperbola. 

B 2 X 

The subnormal = — . 

A 2 

When the focus F (fig. 34) is the origin 

csin. (Po — cr sin. ( p cy 

cotan. * = ' z=z - z= ~. 

A— c cos. t Po B 2 B 2 

When the focus F' is the origin 

c Vo 
cotan. a' =z -~- = — cotan. e 



Corollary. The lines F3I, F' 31, (fig. 52.) drawn from the 
foci to any point 31 of the ellipse make equal angles FMt, 
F t 31 1, with the tangent at the point 31. Hence a tangent 
may be drawn to the ellipse by bisecting the angle F3IF X by 
the line T31t, which will be the tangent required. 

3. The tangent, normal, &c. of the hyperbola are found 
from those of the ellipse by changing the sign of B 2 . 
We hence find for PT (fig. 51.) in this case 

A 2 
x o 

AT — AP — PT=-—, 

x o 
so that for the asymptote we have 

x = oo , A T = 0, 

cotan. s = — oo, «:=t — <p = 0, r = y 

But, in this case, we have by_cor. 1, of the hyperbola, B. I. § 98, 

A 

r = oo , cos. <p = cos. ^ = -p= 



<§> 131.] CONTACT. 229 

Tangent and normal to parabola. 

and the lines EAE' xi E 1 AE' (fig. 36.) are asymptotes of 
the hyperbola. 

4. Find the tangent, normal, &c. of the parabola, whose 
parameter is 4 p. 

a 2 » . y n 

Ans. tang, r = — --, tang, v = — ^-h_. 

The equation of the tangent is 

y y = 2p(x + x ), 
that of the normal is 

*py + yo x = C2p + x o)!/o- 

The subtangent == 2 £ , 

the subnormal = 2p, 

and when the focus is the origin 

sin. tp 2 sin. Ay cos. A <p 

cotan. b = — = £-| — s-^isr — cotan.A«> 

1 — cos. 9 2 cos. J y * 

e = — Jy =: t — 9 

*= Jy; 

that is, if (fig. 39.) M T is the tangent, we have 

MFP = FMQ — cp — 'i FMT, 

so that a tangent may be drawn to the parabola by bisecting 
the angle FMQ. 

5. Find the tangent, normal, &c. of the cycloid. 

Solution. Taking q as the independent variable, wejhave 
by (549) and (566) 

20 



230 DIFFERENTIAL CALCULUS. [b. II. CH. IX. 

Tangent and normal to cycloid and spirals. 

d c . y sin. & ! A 

tang. T = d cx y — - — - = - = cotan. * e 

6 czy d c .x 1 — cos.0 J 

But if (fig. 41.) JfZ? is joined, the angle MBX is measured 
by a semicircumference diminished by half the arc MB, that 
is, 

MBX — 7v — I MCB — n — ^^ — v, 

so that MB is the normal to the cycloid, and MT, which is 
drawn perpendicular to MB, is the tangent. 

The subtangent = 2 R sin. 3 J & sec. £ 5 
the subnormal = .R sin. & = iJ — x 
the tangent = 2 .R sin. 2 \ & sec. J 
the normal =2 R cos. J 6, 

6. Find the tangent of the spirals of equation (133). 

Ans. For the tangent 

n 
cotan. £ = — . 
<Po 

7. Find the tangent of the logarithmic spiral. 

Ans. If the logarithms of equation (136) are taken in the 
system of which the base is a, (136) converted into Neperian 
logarithms is 

log. r = log. a . <p 

and we have, for the cotangent, 

cotan. * == log. a. 



§ 133.] CONTACT. 231 

Order of contact of straight line. 

8. Find the tangent and asymptote of the hyperbolic spiral. 
Ans. For the tangent 

1 
cotan. * z= 

9>o 
dist. of tang, from origin 



which for the asymptote become 

<p = 0, * = r — y = 0, r — (f Q — 0, 
dist. of asymp. from origin =z 2 n R, 
so that the asymptote is parallel to the polar axis. 

132. Corollary. The straight line is completely de- 
termined by the condition, that the first differential co- 
efficient of its ordinate is equal to that of the curve at 
the point of intersection ; and, therefore, the tangent 
has usually only a contact of the first order with a 
given curve ; so that, by art. 121, the tangent does not 
usually cross the curve at the point of contact. 

133. Corollary. A point of one curve may be placed 
upon a point of another, and the two curves turned 
around upon this common pivot until their tangents 
coincide ; in this position, the two curves have evident- 
ly a contact' of the first order. If now one of the curves 
is everywhere of the same curvature, that is, if it is 
a circle, the contact will remain of the same order, 
whichever of its points is brought to the point of con- 
tact j but if it is any other curve, a point of it can 



232 DIFFERENTIAL CALCULUS. [b. II. CH. IX. 

Element of curve. 

usually be found, which, brought to the point of contact, 
will elevate the order of contact to the next higher 
order. 



134. Corollary. By changing the dimensions of one 
of the curves, the order of contact can usually be in- 
creased by unity, for each of the constants which enter 
into the equation of this curve, and upon which its 
dimensions depend ; for each of these constants, regard- 
ed as an unknown quantity, may be determined so as 
to satisfy a new equation of those (545) or (547) upon 
which the order of contact depends. 

Thus, a circle can usually be found, which has a contact of 
the second order, with a given curve at a given point ; a 
parabola, a cycloid, or a spiral of the form (133), (135), or 
(136), which has a contact of the third order; an ellipse or 
hyperbola, which has a contact of the fourth order, &c. 

135. The differential of the arc of a curve is called 
its element. 

Thus, if s denotes the arc, d s is the element of the curve. 
Hence in (fig. 53.) if 

d x = PP< = MN, 
we have then, by regarding MM' as a straight line, 

dy — M'N, ds = MM' = s/{dx* + rfy*,) (570) 
so that if x is the independent variable 

rf, S = V[l+Ky) 2 ] (571) 



§ 137.] CONTACT. 233 

Element of curve. 

and if s is the independent variable, using the notation of art. 
129, 

i = V[(4..*)» + (4..y) a ]. (572) 



136. Corollary. In the same way, if the radii vectores 
AM, AM' are drawn, and the arc MR described with R as a 
radius, we have 

MAM' = d<p 9 M'R = AM' — AM=dr, MR=zrd 9 , (573) 
and if MRM' is regarded as a right triangle 

ds = y/(dr2 -\-r 2 d<P 2 ), (574) 

so that if y is regarded as the independent variable 

d c s=V[(d c r)2 + r2]. (575) 

137. Corollary. The triangle M'MN gives, by (560), x 
being regarded as the independent variable, when none is ex- 
pressed in the formula 

M'MN= r 
C0S * r = Ts = -I-s = dc ' 8X = ^ V (576) 

sin -=K=|f = ^=-- cos -* < 577 > 

and by (564) and (565) 

tug.=^ = y4.,.= ' =^=_-£- (578) 
d c . y . •-." d c . s y sin.* cos." v ' 

normal = yd c s = ^— = -^- = -^-. (579) 

a c . ff z cos.t sin. v v ' 



234 DIFFERENTIAL CALCULUS. [b. II. CH.lX. 

Contact of surfaces. 

138. Corollary. The triangle MRM 1 gives 

31 MR — a (580) 

C0S -'=Ts = d7s = d " r (581) 

<p being regarded as the independent variable, when none is 
expressed in the formula. 



139. Two surfaces which intersect at a given point 
are said to be in contact, when their sections, found by 
any plane passing through this point, are in contact. 

140. Problem. To find the algebraic conditions that 
two surfaces are in contact at a given point. 

Solution. Let the surfaces be referred to rectangular co- 
ordinates, and let x , y , z be the coordinates of the point 
of contact. Then the sections of the two surfaces, made by 
any plane, are found by equations [321-323], and since 
these sections are in contact, the values of d c y lf found on the 
hypothesis that % l is the independent variable, must be equal 
for the two surfaces at the point of contact. Hence the values 
of d c x , d c y , d c z , found by differentiating the equations (321 
-323] must also be equal for the two surfaces, as well as 



d c y n _ dy 
d c x d x 

d„ z n d z, 



= <*«•, y„ (583) 



.1 ; S- = ^ = d c .,z . (584) 



§ 142.] CONTACT. 235 

Tangent plane. 

141. Corollary. If one of the surfaces is the tangent plane, 
its equation, since it passes through the point of contact, is 
by (197) 

cos. « (z— x ) + cos. p (y — y Q ) + cos. Y (z—z ) = 0, (585) 

from which we find by differentiation 

, COS. a 

d -* z =—c^7 = d <- z °' < o87 > 

which substituted in (585), divided by cos. a, give, for the 
equation of the tangent plane, 

(*-*•>- Fr - Fir = °- < 588 > 

142. Corollary. If all the terms of the equation of the 
given surface are transposed to the first member, and this first 
member, which is a function of x, y, z, represented by V, and 
if V becomes V Q at the point of contact, the equation of the 
surface is 

V — 0. (589) 

The differential of this equation being taken on the hypothesis 
that z is constant, in order to find d e . x y Q , gives 

d, x V .dx + d c . y V . dy Q = 0, (590) 

whence, by (586), 

'■''" dx d,. s V cos.? v ; 

or _ _L_ = £?LJ! = £iLl. (592 ) 

d c . x y cos.« d c . z V Q 



236 DIFFERENTIAL CALCULUS. [b. II. CH. IX. 

Tangent plane. 

In the same way we find 

1 cos. r d c . z V 






d c . s z Q ' cos.« d c . x V ' 



(593) 



and these values, substituted in (5S8), give, by freeing from 
fractions, 

4..F. (z-* )-K. y F (y-y„)-K..T. (*-*„) = «• (594) 

This equation of the tangent plane compared with the com- 
plete differential of (589), which is 

d c . s V o dx o +d c . y V o dy o +d c . 2 V o dz =0 (595) 

shows that the equation of the tangent plane may be 
obtained from that of the surface by changing in the 
complete differential of the equation of the surface re- 
ferred to the point of contact dx , dy , dz respectively 
into(# — # ), {y — y Q ), (z — z ). 

143. Corollary. The sum of the squares of (592) and (593) 
increased by unity is 

cos.^+ C os^+cos^y_( d c . x V Q f+(d c . v V f+(d c . z V f 
c^« d^j 2 ( } 

or by (47) and putting 

L =V[(d e -,V )2 + (d c . y V )2 J^(d c . z V o y*] (597) 

cos.«=%^o, d c .,V = L cos.a (598) 

cos. p^tklXsL, d c . v V = L cos. ? (599) 

cos. Y = M%L 9 d c . z V = L cos. y. (600) 



COS 


. a 




cos 


•/5 




cos 


. Y 


X 

d c . x 


X 


= 


y — 

d c . y 


#0 

Vo 


— 


z 


z o 
V 



$ 146.] CONTACT. 237 

Normal to surface. 

144. The perpendicular to the tangent plane drawn 
through the point of contact, is called the normal to 
the surface. 



145. Corollary. The angles «, p, y are, by (128), the angles 
which the normal makes with the axes; hence, by (124) or 
(598-600), the equations of the normal are 

(601) 

146. Examples. 

1. Find the equations of the tangent plane and of the nor- 
mal to the sphere. 

Solution. If the sphere is that of equation (62), we have 

V— Z 2 +y2 + Z 2 __#2 = 

d c . z V = 2z 
d c . y V = 2y 

d c , z V = 2z 

Z = 2 X /K + y§ + zg)r=2i2 

^o „ Vo %o 

COS. « = g", COS. /S = g-, COS. Y == -g-. 

The equation of the tangent plane is, by reduction, 



238 DIFFERENTIAL CALCULUS. [b. II. CH. IX. 

Tangent and normal to ellipsoid. 
The equations of the normal are, by reduction, 
— — !- — — 

x o y o % o 

so that by (128) it passes through the origin and is radius. 

2. Find the equations of the tangent plane and of the normal 
to the ellipsoid of equation (335). 

Arts. The equation of the tangent plane is, by reduction, 

x o x I y$y I z o z i q 

The equations of the normal are 

3. Find the equations of the tangent plane to the cone of 
equation (364). 

Ans. The equation of the tangent plane is 

x ° x i y°y.._ z ° z — 
A 2 -r B 2 c*-~ ' 

so that it passes through the origin. 

4. Find the equations of the tangent plane and of the nor- 
mal to the cylinder of equation (375). 

Ans. The equation of the tangent plane is 

By Q y + Cz Q z + M=^ 

so that it is perpendicular to the plane of y z. The equations 
of the normal are 

x = x 

Cz Q y — By z = (C—B) y z . 



§ 146.] CONTACT. 239 

Tangent and normal to paraboloid. 

5. Find the equations of the tangent plane and of the nor- 
mal to the paraboloid of equation (3/6). 

Ans. The equation of the tangent plane is 

2By y + 2Cz z + H(x + :r ) = 0. 
Those of the normal are 

2Byl(z — x ) = H(y — y ) 
Cz y — By z = (C—B)y Q z . 

6. Find the equations of the tangent plane and of the nor- 
mal to the cylinder, of which the base is a parabola, and the 
equation is (384). 

Ans. Put H cos. a -\- I sin. « — — £p C, and omitting the 
numbers below the letters in (384) ; the equation of the tan- 
gent plane is 

z z = 2p(x-{-x ), 

so that it is perpendicular to the plane of x y. The equations 
of the normal are 

V = I/o 
z (z — *o) == 2p( z c — z)- 



240 DIFFERENTIAL CALCULUS. [b. II. CH. X. 

Radius of curvature. 



CHAPTER X. 



CURVATURE. 



147. The circle, which has the contact of the second 
order with a curve at a given point, coincides more 
nearly with the curve at that point than any other 
circle, and its curvature is therefore adopted as the 
measure of the curvature of the given curve at that 
point. It is hence called the circle of curvature, and 
its centre and radius are called, respectively, the centre 
and radius of curvature. 

148. Problem. To find the radius of curvature of a 
given curve at any point. 

Solution. Let q be the required radius of curvature, * the 
angle which the normal to the given curve makes with the 
axis of x, v ' the corresponding angle for the circle. Then, 
at the point x Q , y Q of contact, we have 

"o = V- ( 603 ) 

But if s, s' are the arcs of the given curve, and of the 
circle, we have, by (576), 

ds = cosec. v. d x, d s' = cosec. v.' dx, 
so that at the point of contact 

ds =ds '. (604) 



<§> 150.] CURVATURE. 241 

Radius of curvature. 

But it is evident from (574) that, since the radius of the 
circle is constant, we have 

ds = ds ' — g d* '. (605) 

But the differential of (561) gives 

dr = sm^r d 2 c . a ;!/ dx o , dr> Q = sm. / v' d 2 c . !e y' Q dx of (606) 

and since the contact is of the second order 

d lVo = d ly'o> 

so that by (603) and (606) 

dr Q = dr , (607) 

which substituted in (605) gives 

ds Q =Qdv , (608) 

or omitting the cyphers below the letters, 

ds d c s , ,„«.,* v 

Q = —=-f-= d c . v s. 609) 

dv d c v v ' 

149. Corollary. Equations (609), (571), (576), (577), (606), 
and (565) give 

t = ^4^- = (^ = tt + ^.3r)-]* (610) 

sm.2vd 2 c . s y d\. s y d\. s y v ' 

( sec t)b N* ds 

Q = -«^=^dl^ = d;= d <'* S ' < 611 > 

150. Corollary. When the equation of the curve is given 
in polar coordinates, the radius of curvature may be found by 
means of equations (557) and (558). For these give 

^ = -^^-(~) (612) 

21 



242 DIFFERENTIAL CALCULUS. [fi. II. CH. X. 

Radius of curyature. 

d r = dq>-{-ds— d(f— sin. 2 *d. (— -P-I (613) 

i = ^ = d^_^ d jd^Y 

q as as as \rd<pj 

151. Examples. 
1. Find the radius of curvature of the ellipse. 
Solution. Equation (69) of the ellipse gives 





B 2 x 




u cxy — 


A 2 y 




d*. x y = - 


B± 

A 2 y 3 




(A*y 2 +B±x 2 ) 2 A 2 


iVs _ 


A 2 B 2 



^ B± B± (A 2 sin. 2 r+S^cos^r)! ' 

This value of the radius of curvature is the same also for the 
hyperbola. 

2. Find the radius of curvature of the parabola of equation 
of B. I. § 180. 

Ans. y + W ^ N* = 2p 
4 p 2 4p 2 sin. 3 T ' 

3. Find the radius of curvature of the cycloid. 

Ans. 4 R sin. J d = 4 JR cos. * — 2 N. 

4. Find the radius of curvature of the spiral of Archi- 
medes. 



Ans. 



2(2 ;* +<*>*) ' 



<§> 154.] CURVATURE. 243 

Evolute and involute. 

5. Find the radius of curvature of the logarithmic spiral, of 
which the equation is given in Example 7, § 133. 
Ans. a? a/[1 + (log. a)*]. 

152. Problem. To find the centre of curvature of a 
given curvature. 

Solution. Let x 1) y l be the coordinates of the centre of 
curvature corresponding to the point of contact x, y of the 
given curve, and, by B. I. §83, x x — x is the projection of the 
radius of curvature upon the axis of x. 

But, by B. I. § 85, this projection is also expressed by 

Q cos. ,, = x L — x, (615) 

whence z 1 = x -J- Q cos. »•. (616) 

In the same way we find 

y x — y + Q sin. v. (617) 

153. Corollary. If the two coordinates of the given 
curve are eliminated between the three equations (616), 
(617), and the given equation of the curve, the result- 
ing equation, containing only the coordinates of the 
centre of curvature, is the equation of the curve upon 
which the centre of curvature is situated. This curve 
is called the evolute of the given curve, for a reason 
which will soon be given. The given curve is called 
the involute of its evolute. 

154. Corollary. The differentials of (616) and (617) are 

dx x — dx -f- do. cos. v — s in. r. q d v (618) 

dy x = dy -\- dQ. sin. v -|_ cos. \ Q d*. (619) 



244 DIFFERENTIAL CALCULUS. [b.II. CH.X. 



Evolute and involute. 


But by (576), (577), and (609) 




sin. v.Qd v = dx 


(620) 


cos. v. q dv '•==. — dy 


(621) 


which, substituted in (618) and (619), give 




dx 1 = dQ. cos. v 


(622) 


dy x = dq. sin. v . 


(623) 



155. Corollary. If ? x and v x are the angles, which the 
tangent and normal to the evolute make with the axis of x, 
we have, by (622) and (623), 

tang. r i = — cot. v x = ~^- = tang, v (624) 

ct x « 

whence 

f t =v= in + *, y x — ^7t-\-v=nJ r r ) (625) 

so that the normal to the involute coincides with the 
tangent to the evolute. 

156. Corollary. If s x is the arc of the evolute, we have, 
by (622) and (623), 

ds x =sf[{dx x y + (dy i y] = ±dQ i (626) 

so that the arc of the evolute increases at the same rate 
that the radius of curvature of the involute increases 
or decreases. Hence 

JS 1 =;kJQ. (627) 

157. Corollary. If CMM' (fig. 54.) is the involute, C[M X M[ 
the evolute, and if MM X) M' M[ are tangent to the evolute, 
and consequently normal to the involute, we have 



§ 160.] CURVATURE. 245 

Evolutes of different orders. 

e = MM lt <?' = 31' M' x 
j Q = 31' 31 ; — 3131 ± = jS— M x M± t 
so that 31' M; = 3I31 1 + 31 1 31[. 

Hence if a string were wound around the evolute of 
such a length, that, when drawn tight at Mj in the 
direction of the tangent M31 1 , it would reach to M, 
it would, when unwound and drawn tight at M{, reach 
to 31', and its extremity would, in the process of un- 
winding, describe the involute. The names of these 
curves are derived from this property. 

158. Corollary. If q x is the radius of curvature of the 
evolute, we have, from (Oil), (625), and (626), 

d s , do 



159. Corollary. The evolute of the evolute is called 
the second evolute, and the evolute of the second evo- 
lute the third evolute, and so on. 

If, then, Q n is the radius of curvature of the n th evolute, and 
\ the angle, which the tangent to this evolute makes with the 
axis of x, (625) and (628) give 

e.-=-±AT« (629) 

Tn = * + in?v. (630) 

160. Scholium. No more natural system of coordi- 
nates of a curve could probably be devised than its 
radius of curvature, q, and the angle, t ; which its di- 
rection makes with a given direction. A curve is 

21* 



246 DIFFERENTIAL CALCULUS. [B. II. CH. X. 

-Evolute. 

readily referred to these coordinates by the equations 
already given ; and from its equation referred to these 
coordinates, the corresponding equation of either of its 
evolutes is readily obtained by means of (629) and 
(630). 

161. Examples. 



1. Find the evolute of the ellipse. 

Ans. The equation referred to the coordinates of § 
160 is 

_ 3 A 2 B 2 (A *— IP) sin . 2t 
q ~ {A^ cos. 2 t + B 2 sin. 2 7)f ' 

2. Find the evolute of the parabola of equation of § 160. 

Ans. Its equation is 

6 p sin. t 

Q == 'a • 

■ COS. 4 T 

3. Find the w th evolute of the cycloid. 

Ans. Its equation is 

q = 4 R cos. r, 
so that it is a cycloid precisely equal to the given cycloid. 

4. Find the n th evolute of the logarithmic spiral. 

Ans. It is a logarithmic spiral. 

5. Find the evolute of the curve of which the equation re- 
sults from eliminating 9 between the two equations 

y — R sin. 9 — R <P cos. 9 

x = R cos. <p + R yjsin. <p. 



§ 161.] CURVATURE. 247 

Involute of circle. 

Solution. We have 

d x = R(pd<p cos. <p 
dy = R(pd<p sin. y 
tang, t = tang. y , T = y 
d s = JRt dx 

Q = R r 
Q ± — d c . r Q == R, 

that is, the radius of curvature of the evolute is constant, and 
the evolute is therefore a circle. 



248 DIFFERENTIAL CALCULUS. [b. II. CH. XI. 

Points of stopping. 



CHAPTER XI. 



SINGULAR POINTS. 



162. Those points of a curve, which present any 
peculiarity as to curvature or discontinuity, are called 
singular points. 

163. Whenever a function is discontinuous, the cor- 
responding curve found, as in § 5, is also generally dis- 
continuous. 

Thus if f. a; is a function of x, which is imaginary for all 
values of x less than 

x = a — AP (fig. 55.) ; 

for all values contained 

between x = a' = AP' and x = a" =z AP" > 
between x =z a'" = AP'" and x z= a IV = AP ly , 

and for all values greater than 

x = a v =z AP" ; 

and is continuous for all values of x 

between x = a z=z AP and x =z a' = AP', 
between x =i a" — AP'' and x = a!" — AP'", 

and between x = a IT = AP IT and a; = a v = AP T ; 



<§> 164.] SINGULAR POINTS. 249 

Points of stopping. 

its locus is composed of the different portions MM' ', M"M"\ 
and M iy M r . 

If, for instance, this function were such as to have always 
the same value 

b = PM — P'N' = PW, &c. 
wherever it was not imaginary, the locus of 

would be the portions 31N', N"N ! " y and N"N Y , drawn par- 
allel to the axis of x. 

164. Examples. 

1. Construct the locus of the equation 

y — b — [log. (x — a)]- 1 

in the vicinity of the point at which it stops j and find its tan- 
gent at this point. 

Solution. The logarithm of a negative number is imaginary, 
and therefore the value of y is imaginary as long as x is less 
than a ; but when x — a, we have 

y — b= (log. 0)- 1 z= oo- 1 =i 

V — ?>> 

so that the point M (fig. 56.), for which 

AP = a } PM = b 

is the point at which the curve stops. At this point we have, 
by § 108, 

tang. rz=z d c .y = — [log. (x — «)]~ 2 (x — a)~ 1 = co 

t— Jtt, 

so that PM is the tangent to the curve at the point M. 



250 DIFFERENTIAL CALCULUS. [b.II. CH.XI. 

Points of stopping. 

* The remainder of the curve near the point M is constructed 
by finding different values of y for different values of x nearly 
equal to a, and drawing the curve through the points M, M 1 , 
M", &c, thus determined. The figure 56 has been constructed 
for the case in which 

a = 2, b — I, 

and, for the present example, extends to 

x — AP" = 2.135, y == 0o, t = 118° 26'. 

2. Construct the locus of the equation 

y — b — (x — a) [log. (x — a)]- 1 

in the vicinity of the point at which it stops ; and find its tan- 
gent at this point. 

Ans. This locus is, for the present example, represented 
in fig. 57, from 

x = — oo to x zzz a -{- 0*5 =r AP'. 

The point where the curve stops corresponds to 

x = a z= AP, where r = 0, 

so that 31 T parallel to AX is the tangent. 

3. Construct the locus of the equation 

y — b — (x — a—\) [log. (x — a)]- 1 

and find the tangent at the point where it stops. 

Ans. This locus is represented in fig. 58. The point 
where it stops corresponds to 

x =z a — AP, where r = J n, 
so that PM is the tangent. 



§ 164.] SINGULAR POINTS. 251 

Points of stopping. 

4. Construct the locus of the equation 

y — b = (x — a) log. (x — a) 
and find the tangent at the point where it stops. 

Ans. This locus is represented in fig. 59. The point 
where it stops corresponds to 

x = a == AP, where r — J w 
so that PM is the tangent. 

5. Construct the locus of the equation 

y — b — (x — a) 2 log. (x — a) 

in the vicinity of the point where it stops, and find the tangent 

at this point. 

Ans. This locus is represented in fig. 60, which, for the 
present example, extends 

from x =z — oo 

to x=AP' = a + 0-223, where y=b— 0-075, r = 155° 57'. 

The curve stops at the point corresponding to 

x — AP = a, where r = 0, 

so that 3IT, parallel to AX, is the tangent. 

6. Construct the locus of the equation 

y — b — (x — a) [log. (x — a)Y 

v, 

in the vicinity of the point where it stops, and find its tangent 
at this point. 

Ans. This locus is represented in fig. 61, which, for the 
present example, extends 

from x = — oo to x = AP' = a + 0-368 ; 



252 DIFFERENTIAL CALCULUS. [b. II. CH. XI. 

Points of stopping. 

it stops at the point coresponding to 

x = AP = a, where T = £ rc i 
so that PM is the tangent at this point. 

7. Construct the locus of the equation 

y — b = (x — a) 2 [log. (x — a)] 2 

in the vicinity of the point where it stops, and find the tangent 
at this point. 

Ans. This locus is represented in fig. 62, which, for the 
present example, extends 

from x = — oo, to x = AP' = a + 0-38 j 
it stops at the point corresponding to 

x ±j= AP — a, where r = 0, 
so that MT, parallel to AX, is the tangent. 

8. Construct the locus of the equation 

y = log. (x+l) + zlog.a; 
and find the tangent at the point where it stops. 

Ans. This locus is represented in fig. 63 ; it stops at the 
origin where the axis of y is the tangent. 

9. Construct the locus of the equation 

yz=mx log. x-\-n{a — x) log. (a — x) 
and find the tangents at the points where it stops. 
Ans. This locus stops at the points where 
x — and x — a : 



§ 164.] SINGULAR POINTS. 253 

Points of stopping. 

at which points the values of y are, respectively, 
y = na log. 0, and y — ma log. a ; 
and the tangents are parallel to the axis of y. 
Figure 64 represents this locus when 

= 1, m = 2, n = 3, 
and figure 65 represents it when 

«=1, m = 2, ?i = — 3. 

10. Construct the locus of the equation 

y=zmx 2 log. x -(- n (a — x) log. (0 — x) 
and find the tangents at the points where it stops. 
Ans. This locus stops at the points where 
x = and x = ; 
at the first of which points 

tang, t = — rc log. (0 — x) — n, 
and at the second 

Figure 66 represents this curve when 

a = 1, m = 2, n =. 3; 
and figure 67 represents it when 

= 1, m =1 2, w = — 3. 

11. Construct the locus of the equation 

#=/i- *(/•*)" log./.* 

in which /. x is a given function of z ; and find the points 
where it stops. 

22 



254 DIFFERENTIAL CALCULUS. [b. II. CH. XI. 

Points of stopping. 

Ans. It stops when f. x becomes imaginary, or when it be- 
comes negative. 

Figure 68 represents this locus when 

f x .x — n~\, f.x — x 2 — x, 
in which case it extends 

from x =z — oo to x =z 0, 
where it stops, and extends again 

from x = 1 = AP to x = oo. 

The tangents at each of the points where it stops is parallel to 
the axis of x. 

Figure 69 represents this locus when 

n = 1, f v x = x, f.x = z 2 — Xy 

so that the points of stopping are the same as in figure 68. 
But the tangent at the point A is the axis of x, while that at 
the point P is parallel to the axis of y. 

Figure 70 represents this locus when 

n — %> /i-z=l, /. x = x 2 — x, 

so that the points of stopping are the same as in figure 68 ; 
but the tangent at each point is the axis of x. 

Figure 71 represents this locus when 

f v x = n = 1, f.x = x—-x 2 , 

so that the points of stopping and the tangents are the same 
as in figure 68 ; but the curve extends from one point to the 
other. 

Figure 72 represents this locus when 

»S=1, f lt Z=ZX t f.Z=ZX—z2 f 



<§> 164] SINGULAR POINTS. 255 

Points of stopping. 

so that the points of stopping and the tangents are the same 
as in figure 69 ; but the curve extends from one point to the 
other. 

Figure 73 represents this locus when 

n = 2 t f 1 .x=l, f.x=zz— x 2 , 

so that the points of stopping and the tangents are the same 
as in figure 70 ; but the curve extends from one point to the 
other. 

Figure 74 represents this locus when 

n = l,f 1 .z = (10z + l)-i~ 

f.z = z(x — l){x—2)(x — 4) (x - 5) 

in which case it extends 

from x = to x = 1 = AP , where it stops, 

from x = 2 = AP 2 to x = 4 = AP \ where it stops, 

from z = 5 = AP to z ■=. oo. 

The tangent at each point where it stops is parallel to the axis 
of y. 

Figure 75 represents this locus when 

n = l.f 1 .z = (Wx + \)-i . 
/ x = - x (I _ 1 ) (x - 2) (x - 4) (x - 5), 

in which case it extends 

from x = — oc to x = 0, where it stops, 

from x = l = AP 1 to x = 2 =AP 2 , where it stops, 

from x = 4 = AP± to x = 5 = AP 5 , where it stops ; 

the points of stopping and the tangents are the same as in 
figure 74. 



256 DIFFERENTIAL CALCULUS. [b. II. CH. XI. 

Points of stopping. 

Figure 76 represents this locus when 

« = l,/ 1 . 2; = J(a; + ])^( 2: _2)( x __3)(x~-4)(2 2 :+3)-2 

/* = x(x_l) (x-2) (i — 4) (x — 5); 

in which case it extends as in figure 74, and the tangents at 
the points P 1 and P 5 are parallel to the axis of y, but the 
axis of % is the tangent at the points A, P 2 , and P 4 . 

Figure 77 represents this locus when 

n= 1, / r x = 4 (x+ 1) x* (x— 2) (s— 3) (x— 4) (10x+l)- 4 

/.x = -x(x-l)(x-2)(x-4)(x — 5), 

in which case, it extends as in figure 75, and the tangents are 
as in figure 76. 

Figure 78 reprents this locus when 

»=!, f 1 .x=(10x + l)-i 

f.x = x (x-1) (x — 2) (x— 3) (x — 4) (x — 5), 

in which case it extends 

from x = — oo to x = 0, where it stops, 

from z=zl=AP 1 to x = 2z=:^4P 2 , where it stops, 

from x = 3 = AP 3 to x = 4 = AP±, where it stops, 

from x = 5 = ^lP 5 to x = oo; 

the tangent at each point where it stops is parallel to the axis 
of y. 

Figure 79 represents this locus when 

71=1, ^.X^^IOX+I)-* 

/.* = — x(x — l)(x — 2)(z — 3)(x — 4)(x_ 5), 
in which case it extends 



§ 164.] SINGULAJR POINTS. 257 

Points of stopping. 

from x = to x = 1 = AP 1} where it stops, 

from 2=2 = AP 2 to z = 3 = AP 3 , where it stops, 

from x = 4 c = AP 4: to z = 5 = .4P 5 , where it stops ; 

the points of stopping and the tangents are the same as in 
figure 78. 

Figure 80 represents this locus when 

n = l ) f 1 .x=i(x+l)x2(x-2)(x-3)(z-4)(2x + 3)-2 

/.z = z(z_l)(z-2)(z-3)(z--4)(z-5), 

in which case it extends as- in figure 78, and the tangents at 
the points P x and P 5 are parallel to the axis of y\ but the 
axis of z is the tangent at the points A, P 2 , P 3 , and P 4 . 

Figure 81 represents this locus when 
n = l,/ 1 .i = 4(i + l)z2(x— 2)(z — 3)(z — 4)(10z+l)-« 
/.z = — z(z — l)(z-2)(z— 3)(z — 4)(z—5), 

in which case it extends as in figure 79, and the tangents are 
as in figure 80. 

Figure 82 represents this locus when 
n= 1, f 1 .x = ±x 
/.z = — 6z(z_ 1)2 (z — 2), 
in which case it extends from 

i = 0toz = 2 = ^4P 2 . 

Figure~83 represents this locus when 

» = i; / 1 z = iz(« — i)-i 

/.z = — 6x (z — 1)2 (z — 2), 

in which case it extends as in figure 82. 
22* 



258 DIFFERENTIAL CALCULUS. [b. II. CH. XI. 

Points of stopping. 
Figure 84 represennts this locus when 

f.x = x-j- \/x, 

in which case the portion AM of the curve, which corresponds 
to the positive value of the radical, extends 

from x = to x z=i oo j 

the portion P 1 M 1 , which corresponds to the negative value 
of the radical extends 

from x = 1 = AP x to x = oo . 

Figure 85 represents this locus when 
n = 2, f r x=l 
f.x — x-\-\/x, 
in which case the portions extend as in figure 84. 

Figure 86 represents this locus when 

n = 0, f v x = (x 2 — x) 
f.x — x-\- s/x, 
in which case the portions extend as in figure 84, 

Figure 87 represents this locus when 

71— 1, f v Xz=. \0g. f.X 

fx—x-\- */x, 
in which case the portions extend as in figure 84. 
Figure 88 represents this locus when 
n=z 1, f 1 .x=l 

fx=z{x + */x)*, 



§ 164.] SINGULAR POINTS. 259 

Points of stopping. 

in which case each portion extends from 
x — to x r= oo . 
Figure 89 represents this locus when 

n = l t f v x=(x + */x)-i 

f.X=(x + *Sx)2 

in which case the portions extend as in figure 88. 
Figure 90 represents this locus when 

n— 1, f x .x — log./, x 
f.x = (x + *Sx)2, 
in- which case the portions extend as in figure 88. 

Figure 91 represents this locus when 

n = 0, f 1 .x=(x2 — x)2\og.f.x 
f.x=(x + V*) 2 , 
in which case the portions extend as in figure 88. 
Figures 92 - 99 represent this locus when 

f 1 .x=n=l, /.x = a + V(4— x 2 ), 
in which case it stops at the values 
x = — AP=l— V(4 — a 2 ) andxz=^P" = V(4 — a 2 ). 

The tangents at the points P' and P" are parallel to the axis 
of y. 

In figure 92, a = — 1.5. 

In figure 93, a = — 1 . 

In figure 94, a =z — 0.5. 

In figure 95, a =. . 



260 DIFFERENTIAL CALCULUS. [b. II. CH. XI. 

Points of stopping. 

In figure 96, a r= 0.5. 

In figure 97, a = 1. 

In figure 98, a z=. 1.5. 

In figure 99, a =. 2. 

12. Construct the locus of the equation 

y=f'X + mf v x 

when m is infinitely small and f x , x a junction, which is not 
infinite while x is finite, but is alternately real and imaginary. 

Solution. Since m is infinitely small, the part m f x .x may 
be neglected when f t .x is real, but when f x . x is imaginary, 
the value of y is imaginary, so that if figure 100 is the locus 
of equation 

y — f. * ; 

the same figure, with the dotted parts omitted, which corre- 
spond to the imaginary values oif 1 .x ) represents the locus 
of the given equation. 

Thus, the locus of the equation 

y = \/{R 2 — x 2 ) + 0-00000000001 X x n log. x 

differs insensibly from the semicircle BCB 1 (fig. 101.), of 
which R is the radius. But it must be remarked, that, when 
n is zero, the curve is suddenly turned into the form of a hook 
at the points B and B', so as to become tangent to the axis 
of y, assuming a form similar to that of the dotted line, but 
of indefinitely less extent. 

13. Construct the locus of the equation 

r=f.<p + mfi.q. 



<$> 166.] SINGULAR POINTS. 261 

Conjugate points. 

expressed in polar coordinates, and in which m is infinitely 
small, and f x . q finite when real. 

Solution. If MM' M"M '" &c. fig. (102.) represent the 
locus of 

r=-f.q>, 

the same curve with the dotted parts omitted represents the 
given curve, the dotted parts corresponding to the imaginary 
values of f v cp. 

Thus, if /• 9 = -R> 

the curve consists of several successive arcs of the same cir- 
cle. 



165. A conjugate point is one separated entirely from 
the rest of the curve, but included in the same algebraic 
equation. 

A conjugate point is indicated algebraically by the condition 
that coordinates of this point are real, while the coordinates 
of no adjacent point are so. 



166. Examples. 

1. Construct the locus of the equation 

V =f' x + m fi- x > 
in which m is imaginary and f x . x real. 

Solution. If the curve (fig. 100.) is the locus of 
V = f- z> 



262 DIFFERENTIAL CALCULUS. [b. II. CH. XI. 

Conjugate points. 

and if M, M', &c. are the points which correspond to the ab- 
scissas, for which 

the locus of the given equation consists of the series of conju- 
gate points M, M' } &c, without any continuous curve. 

If f. x -= a x -|- b t 

all these points are upon the same straight line. 

2. Construct the locus of the equation 

(y — / *) 2 + (y —/.•*)" = o- 

Solution. The sum of two squares cannot be zero unless 
each square is zero ; so that the given equation is equivalent 
to the two equations 

y — f.x = t y—f 1 .x = 0; 

that is, the coordinates of all the points of the required locus 
satisfy these two equations. 

If, then, App< P"P<» P" P* &c. (fig. 103.) is the locus of 
the equation 

and if AP 1 P I P[P" p\P< P lY P 1 ? &c. is the locus of the 
equation 

the required locus is the series of conjugate points A, P' } P" 9 
P IV , &c, in which these curves intersect. 

Thus the locus of the equation 

(X — fl)2 + (y - 6)2 = 



§ 166.] SINGULAR POINTS. 263 

Conjugate points. 

is the single conjugate point of which the coordinates are a 
and b. 

3. Find the conjugate points of the locus of the equation 

y = f- z +fi- x f 2 - x > 
in which f r x and f 2 . x are sometimes imaginary. 

Solution. \ff v % is imaginary for values of x between a 
and b, and, if f 2 . x vanishes for one or more of the values of 
x contained between a and b ; the given equation is reduced 
for these values of f 2 . x to 

y = /-*> 
so that the corresponding points of the curve are conjugate 
points situated upon the locus of the equation 

9 = /• * 

In the same way, those points of this locus are conjugate 
which correspond to values of x, for which f x . x vanishes, 
while f 2 .x is imaginary. 

Thus the point P , for which x = — 1 is a conjugate point 
upon the axis of x in the curve of figure T6. 

This locus is represented in figure 104, when 

f.X = K /(±-x2) ) f i :X = s/(l-x2),f 2 .X=x\0g.X-l. 

It has four conjugate points, 31 , J/j, 31 , 31^, situated 
upon the circle of which the origin is the centre, and of which 
the radius is 

AP = — 2. 

The common abscissa of two of these points is 

z = — AP> = — 1, 



264 DIFFERENTIAL CALCULUS. [b. II. CH. XL 



Branch. Multiple points. Cusp. 

and of the other two 

x = AP" = 1-763. 

4. Construct the locus of the polar equation of example 13, 
§ 164, when m is imaginary. 

Arts. It represents a series of conjugate points, upon the 
curve of which the equation is 

r—f. T 

These points correspond to the values of <p, which satisfy the 
equation 

A- v = o. 

When f.<p = R, 

the points are all situated upon the circumference of which R 
is the radius, and the origin the centre. 

167. A branch of a curve is a continuous portion of 
it, which extends from one point of discontinuity to 
another. 

When a branch returns into itself, so that its com- 
mencement is the continued curve of its end, it is called 
an oval. 

168. A point through which the curve passes more 
than once, or at which two or more branches termi- 
nate, is called a multiple point. 

A multiple point at which two or more branches stop, 
and have the same tangent, is called a cusp. If a branch 
begins and ends at a point, having but one tangent at 
this point, without being continuous, this point is also 
a cusp. 



<§> 170.] SINGULAR POINTS. 265 

Portions and branches. 

A cusp is said to be of the first kind, when the two 
branches at the point of contact lie upon opposite 
sides of the tangent, as at M (fig. 106.) ; but if the two 
branches lie upon the same side of the tangent, as at 
M (fig. 107), the cusp is said to be of the second 
kind. 

169. In the algebraic consideration of curves, they 
are naturally divided into portions, according to the 
number of ordinates which correspond to the same 
abscissa ; or of radii vectores, which correspond to the 
same angle. 

"The algebraic portions of a curve are not to be confounded 
with the geometric branches; for the same portion may consist 
of several branches, or several different portions may be united 
into one branch. 

Thus the cycloid consists of but one portion, but of an infi- 
nite number of branches ; whereas the circle, the ellipse, and 
the parabola consist of two portions, but only of one branch ; 
and though the hyperbola consists of two portions and two 
branches, yet half of each portion belongs to each branch. 

170. Problem. To find the cusps of a given curve. 

Solution. 1. If a portion M'MM" (fig. 105.) of a curve, 
whose equation is expressed in rectangular coordinates, has a 
cusp at a point M, it is evident that the tangent TM at this 
point must be perpendicular to the axis of x. For if it were 
not so, as in figure 106, there would, for the abscissa AP' 
very near to AP, be the two ordinates P ' M' and P 1 M[, so 
that MM 1 and MM[ would be two different portions of the 
curve, and not the same portion, as we here suppose. 
23 



266 DIFFERENTIAL CALCULUS. [b. II. CH. XI. 

Cusps. 

Moreover, the tangents M' T 1 and M " T", which are in- 
finitely near to M T must evidently be inclined to the axis of 
x } one by an acute angle and the other by an obtuse angle, so 
that 

tang. x—d c .y 

must change its sign at the point M[ by passing 
through infinity if the point M is a cusp, formed by 
two branches of the same portion of the curve ; and 
such a cusp is necessarily of the first kind. 

2. If a cusp is formed at the meeting of two 
branches of different portions, as at M {figs. 106 and 
107) and if the common tangent MT is not perpen- 
dicular to the axis of x ; the ordinates for both por- 
tions, which correspond to the abscissas AP and AP', 
one of which is greater and the other less than AP, 
must be imaginary for one of these abscissas, and real 
for the other. The cusp is of the first kind, as in 
figure 106, if the value of r is greater than MTX 
upon one branch, and less than MTX upon the other 
branch ; but it is of the second kind, as in figure 
107, if the value of r is greater or less than MTX 
upon both branches. 

But if the common tangent is perpendicular to the 
axis of x at M {figs. 108 and 109), the ordinates 
for the two portions must be both increasing, or both 
increasing in proceeding from M. The cusp is of 
the first kind, as in figure 108, when it is the end 
of one branch and the beginning of the other ; but it 



§ 173.] SINGULAR POINTS. 267 

Branches. Oval. 

is of the second when it is the end or the beginning 
of both branches , as in fig. 109. 

171. Problem. To find the points where two por- 
tions of a curve unite in the same branch. 

Solution. The point M (figs. 110 and 111) is one 
of the required points, if the two portions MM' and 
MM[ have a common tangent at this point, while the 
point is not a cusp, but - merely a point where both the 
portions stop. 

172. Corollary. The portions M M' M 2 and 
MM' X M 2 (fig. 112) compose an oval, if at their two 
extremities M and M 2 they unite in a continuous curve 
and have no point of discontinuity between their ex- 
tremities. 

173. When the curve is expressed in polar coordi- 
nates, the analytic portion depends upon the number of 
radii vectores which correspond to the same angle. 
But it must not be overlooked that the same direction 
is determined by angles which differ by any entire mul- 
tiple of four right angles, so that a curve like one of the 
spirals of B. 1, <§> 98, may consist of but one portion, 
although there are an infinite number of radii vectores 
in each direction. 

Multiple points are obtained in any portion, when the 
same value of the radius corresponds to two or more 
angles, which differ by any entire multiple of four right 
angles. 



268 DIFFERENTIAL CALCULUS. [b. II. CH. XI. 

Branches and multiple points. 

174. Examples. 

1. Find the cusps of the cycloid. 

Solution. The cycloid obviously consists of but a single 
portion. If there is a cusp, the tangent at it must, then, as 
in § 170, be perpendicular to the axis of x ; that is, we must 
have, by § 131, example 5, 

cotan. ^ b z=z oc , 
which gives 

J 3 = n Ttj & = 2« n, 
in which n is an integer. 

But this value of & gives, by (131), 

y = 0, 

and since we can never have 

cos. 6 > 1, 

the value of y is never negative, so that there is a cusp at 
each of the points where 

y = 0. 

2. Find the branches of the locus of Example 1, § 164. 

Arts. It consists of two branches, one of which, MM 
(fig. 56.), begins with 

x = a, y = b, 
and extends to 

x = a-{- I, y — — qo . 
The second M x M[ begins with 

a:==a + l, y = oo, 



<§> 174.] SINGULAR POINTS. 269 

Branches and multiple points. 

and extends to 

X = oo , y = b. 

3. Find the branches of the locus of Example 2, § 164. 
Ans. It consists of two branches, one of which MM' (fig. 

57.), begins with 

x = a, y = b, 

and extends to 

x = a -\- 1, y = — oo. 

The second M x M[, begins with 

x— a -\- 1, y = oo, 

and extends to 

£ = oo , y = oo . 

The least value of y in this second branch is found by 
<$ 113, to be 

y = PjilSf; = 2.718, corresponding to z = 4P; = 2.718. 

4. Find the multiple point of the locus of Example 11, 
§ 164, when 

» = 0, f 1 .x=(z*—x) 

f.x — x-\- */x. 
Ans. There is a multiple point when 

x = 1 = AP 2 , y = 0, 

at which point the portion corresponding to the negative value 
of the radical begins, its tangent being P 2 T 2 (fig. 86.), 
drawn parallel to the axis of y, and the portion corresponding 
to the positive value of the radical passes through the same 
point, its tangent being P 2 T 2 \ so drawn that 

P 2 T 2 X—Z4P M, 

23* 



270 



DIFFERENTIAL CALCULUS. J [b.II. CH.XI. 



Branches and multiple points. 



5. Find the multiple points of the locus of Example 11, 
§ 164, when 

n=l, f 1 .z = log.f.z 

f.x = i(x-\- */y>). 

Ans. There are two multiple points; one is atP x (fig. 113.) 
where 

xz= 1, yzrO; 

at which point the branch corresponding to the negative value 
of the radical begins, while the other branch passes through 
it; P x T x is the tangent to the former branch, and the axis 
of x is tangent to the latter branch. The other multiple point 
is at M 2 , where 

x =196, y = 0-45; 
the value of r for the former branch is 
r — 149° 15', 
and that for the latter branch is 

r == 60° 35'. 



6. Find the cusp and the other multiple point of the locus 
of Example 11, § 164, when 

n =z 1, f v z =z I, f. x = (x + ^/x)2. 

Ans. The origin A (fig. 88.) is a cusp of the second kind, 
and the axis of x is the tangent at this point. 

The other multiple point M x corresponds to 

x — 0-328, y == 0-169. 

The values of * at this point are 

T = 69° 29', and r = 6° 30'. 



$ 174.] SINGULAR POINTS. 271 

Branches and multiple points. 

7. Find the branches and the multiple point of the locus of 
Example 11, § 164, when 

n=l, f v z = (x + A/z)-\ f.x = (x + */x)2. 

Ans. The curve consists of but one branch, for the two 
portions unite in one branch at the origin A (fig. 89.) 

The multiple point corresponds to 

x z= 0-544, y = 0-634, 

at which point the two values of * are 

t — 76° 35', r — 124° 13'. 

8. Find the multiple points of the locus of Example 11, 
§ 164, when 

n = 1, /i-z = log./.z 
/. x = (x + s/xy. 

Ans. The origin (fig. 90.) is a cusp of the second kind, the 
tangent at this point being the axis of x. 

31 1 is a multiple point corresponding to 

x = 0-142, y z=z 0-465, 
at which point the two values of * are 

t— 114° 37', t — 21° 40. 

31 2 is a multiple point corresponding to 

x ±= 0-544, y — 0-402, 
at which point the two values of t are 

r — 53° 17', r — 172° 29'. 

9. Find the multiple points of the locus of Example 11, 
§ 164, when 



272 DIFFERENTIAL CALCULUS. [b. II. CH. XI. 

Branches and multiple points. 

n = 0, f x x = (x 2 — x) 2 log./, x 
f.x=(x + s/x) 2 . 

Arts. The origin (fig. 91.) is a cusp of the second kind, the 
tangent at this point being the axis of x. 

M x is a multiple point corresponding to 
z = l, y = 0, 

at which the axis of x is the common tangent to the two 
branches of the curve, and the contact of the two branches is 
of the second order. 

10. Has the locus of Example 11, § 164, any cusp, when 
n=zl,f 1 .x = ix(x— l)-i,/.x = — 6z(z— l) 2 (s— 2)? 
Arts. It has none. 

11. Find the multiple point of the locus of Example 11, 
§ 164, when 

f 1 .x = n=l, /.x = a + V(4-x2). 

Arts. When a is zero, or negative, there is no multiple 
point. 

When a is positive, and less than 

c- 1 == [2-71828]- 1 = 0-3679, 

the curve consists of a single branch without any multiple 
point. 

When a = e- 1 = 0-3679 

the curve consists of three branches, as in (fig. 114), and has 
two cusps of the second kind, correspondsng to 

z = ±2, y = — a = — 0-3679. 



§ 174.] SINGULAR POINTS. 273 

Branches and multiple points. 

When a is greater than er~ 1 and less than ^, the curve con- 
sists of one branch with two multiples, as in figure 115, where 

a = 4, 

and the two multiple points correspond to 

x = ± 1-984, y — — 0-275. 
The values of * at one point are 

t — 102° 38', and % = 97° 45', 
at the other 

*■=: 77° 22', and r — 82° 15'. 
When a — J ~ 05, 

the curve (fig. 96.) has two multiple points at the beginning 
and end of its branch, corresponding to 

x — ± 1937, y — 0. 

The values of * are 

T — 90° , , rrr90°± 45° . 

When a is greater than J and less than 2, the curve consists 
of a single branch, with no multiple points. 

When a = 2, 

the curve (fig. 99.) is an oval. 

12. Construct the locus of Example 11, § 164, when 
f 1 .x = n = l, f. x = a + \/(« 2 — z 2 ). 

Ans. Where a is greater than J, the curve is an oval, as in 
figure 99, where 

a — 2. 

When a = \ = 05, 



274 DIFFERENTIAL CALCULUS. [b. II. CH. XL- 

Branches and multiple points. 

the curve (fig. 116.) consists of a single continuous branch, 
which returns into itself; and it has a multiple point at the 
origin, where the curve has a contact with itself, the common 
tangent being the axis of x. 

When a is less than J and greater than 

c-i = 0-3679, 

the curve consists of a single continuous branch, which re 
turns into itself; and has two multiple points, as in figure 117, 
where 

a = 0-4, 

the two multiple points correspond to 

x = ±0-31, y= — 0-27. 
The values of * at one point are 

t = 144° 53' r = 131° 41', 
and at the other 

r = 35° 7' r = 45° 17'. 
When a = e~* = '0-3679, 

the curve (fig. 118.) consists of two branches and two cusps 
of the second kind, corresponding to 

x = a, y = — a. 

When a is less than er~ l , the curve is an oval, as in figure 
119, where 

a = 0-2. 

13. Construct the locus of the equations of Example 5, 
§ 161, and find its cusp. 

Arts. This locus is represented in figure 120. Its cusp is 
of the first kind, and corresponds to 



■ 



<§> 174.] SINGULAR POINTS. 275 

Branches and multiple points. 

<P = 0, z = 1, y = 0, 
where the axis of z is the tangent. 

14. Construct the locus of the equation 

y 2 — %% 

and find its cusp. 

Ans. This locus is represented in figure 121. Its cusp is 
of the first kind, and is the origin, where the axis of x is the 
tangent. 

15. Construct the locus of the equation 

y 2 == x 4 — x 3 , 
and find its cusp. 

Ans. This curve (fig. 122.) consists of three branches, two 
of which extend from 

x = — oo to x — 0, 
where there is a cusp of the first kind. 
The third branch extends from 

x — 1 to x — oo . 

16. Construct the locus of the equation 

y 2 — x 3 — x*, 

and find its cusp. 

Ans. This locus (fig. 123.) consists of a single branch, 
which has a cusp of the first kind at the origin. 



17. Construct the locus of the equation 
and find its branches. 



y 2 =z x 4 * — x 2 , 



276 



DIFFERENTIAL CALCULUS. [b. II. CH. XI. 



Branches and multiple points. 



Ans. This locus (fig. 124.) consists of two branches, one of 
which extends from 

X Z=Z — oo to x = — 1 , 

and the other from 

x — 1 to x =z oo , 

and a conjugate point, which is the origin. 



18. Construct the locus of the equation 



y 



and find its multiple point. 



x 2 — x±, 



Ans. This locus (fig. 125) consists of one branch, which 
returns into itself, and has a multiple point at the origin, where 
the values of r are 

r = dz 45°. 
19. Construct the locus of the equation 



and find its multiple point. 

Ans. This locus (fig. 126.) consists of a single branch, 
which returns into itself, and has a multiple point at the origin, 
where it has a contact with itself. The tangent at the origin 
is the axis of x. 



20. Construct the locus of the equation 



y 2 — x 3 — x. 



Ans. This locus (fig. 127.) consists of an oval, which|ex- 
tends from 






§ J 74.] SINGULAR POINTS. 277 

Multiple points. 

X = — 1 to x z= 0, 
and a branch which extends from 

x = 1 to x ±t oo . 

21. Construct the locus of the equation 

y 2 z= x 5 — a; 3 , 
and find its cusp. 

Ans. This locus (fig. 128.) consists of a branch, which ex- 
tends from 

x = — 1 to x = 0, 
where there is a cusp^and a branch, which extends from 
x — 1 to x — 00 . 

22. Construct the locus of the equation 

%J 2 = x 2 (1 —x 2 ) 3 , 
and find its multiple points. 

Ans. This locus (fig. 129.) consists of two branches, which 
extend from 

x = — 1 to x == J. 
They cross at the origin, where the values of t are 

t = d= 45°, 
and there are two cusps of the first kind, corresponding to 
z = d= 1. 

23. Construct the locus of the equation 

y 2 — x± (1 — x 2 ) 3 , 
and find its multiple points. 
24 



278 



DIFFERENTIAL CALCULUS. [b. II. CH. XI. 



Multiple points. 



Ans. This locus (fig. 130.) consists of two branches, which 
extend from 

x = — 1 to X = 1. 

There are two cusps corresponding to the two values of x, and 
the origin is also a multiple point, where the two branches are 
in contact. 



24. Construct the locus of the equation 

y* = (2 — x2) (1 — X 2) d — >), 

and find its branches. 

Ans. This locus (fig. 131.) consists offa succession of three 
ovals, which extend respectively 

from x — — \/2 to x — — 1 

from x — — \ZJ to x — s/\ 

from x = 1 to x = s/ 2. 

25. Construct the locus of the polar equation 

r = a -f- sin. m cp, 

and find its multiple points and branches. 

Ans. If m is an integer and a greater than 1, this locus is 
oval, as in figure 133, where 

a = 2, m = 3, 

If m is a fraction and a greater than 1, this locus is a curve, 
which returns into itself after as many revolutions of the radius 
vector as there are integers in the denominator of m. 



Thus, in (fig. 134.), 

a = 2, m 



h 



§ 174.] SINGULAR POINTS. 279 

Multiple points. 

there is a multiple point corresponding to 

q> — or — 360°, r — 2, 
at which point the values of e and * are 

« = r — 75° 58', and ■ = r = 104° 2'. 

In (fig. 135.) a — 2, m — §, 

there are three multiple points corresponding to 
q>=z0° or =360°, <pz=120° or =480°, ^ = 240° or =600°, 

at each of which points the values of r and * are 
r = 2, s = 53° 8', and . = 126° 52'. 

In (fig. 136.) a = 2, m = J, 

there are two multiple points, one of which corresponds to 
<p z= 90° or = 450°, r = 2.5 
6 = 83° 25' and * = 96° 35', 
and the other to <p — 630° or = 990°, r = 1-5. 
a — 79° 7' and * = 100° 53'. 

In (fig. 137.) a = 2, m = f , 

there are four multiple points ; at two of these points we 
have 

q> — 45° or = 765°, <p — 225° or = 585°, 
and at each of these points 

r = 2.5, * = 77°, and * — 103° ; 
at the other two points we have 

<p = 315° or == 1035°, <p — 495° or = 855°, 



280 



DIFFERENTIAL CALCULUS. [b. II. CH. XL 



Multiple points. 



and at each of these points 

r — 1.5, b — 68° 57', and * — 111° 3'. 



In (fig. 138) a — 2, m = J, 

there are three multiple points ; one of which corresponds to 
<P = 0° or == 720°, r — 2 
* — 82° 53' and e — 97° 7' ; 
the second corresponds to 

cp — 180° or == 540°, r — 2.707 
s = 86° 16' and « = 93° 44'; 
the third corresponds to 

9 = 900° or = 1260°, r = 1.293 
b = 82° 13' and * = 97° 47'. 

In (fig. 139.) a = 2, m = f, 

there are nine multiple points ; three of which correspond to 
9 = 0° or =720°, ^=240° or = 960°, y = 480° or =1200°, 
and at each of these points we have 

r = 2, . = 69° 27', and * = 110° 33' ; 
three correspond to 

= 60° or =1140°, 9=180° or =540°, <p—660° or = 1020°, 
and at each of these points, we have 

r == 2.707, * = 78° 55', and * = 101° 5'; 
three correspond to 



§ 174.] SINGULAR POINTS. 281 

Multiple points. 

<*>=300° or = 1380°, <p z= 420° or = 780°, <p=900° or = 1260° 

and at each of these points we have 

r = 1.293, • = 67° 42', and * = 112° 18'. 

If a — 1 and m an uneven integer, or a fraction whose 
numerator is uneven, the origin is a multiple point consisting 
of the union of as many cusps as there are in the integer m, 
or the numerator of the fraction m. 

Thus, in (fig. 140.) a = 1, m = 1, 

the origin is a cusp, and the tangent is the axis of y. 

In (fig. 141.) a — 1, m = 3, 

the origin is an union of three cusps, and the three values of 
t at this point are 

r — 90°, r — 210°, r — 330°. 

In (fig. 142.) a = 1, m = 5, 

the origin is an union of five cusps, and the five values of * at 
this point are 

T == 54°, r = 126°, t -= 198°, 

r = 270°, r t= 342°. 

In (fig. 143.) a = 1, m = J, 

the origin is a cusp, and the tangent at this point is the axis of 
x. There is a multiple point corresponding to 

9 = 0° or z= 360°, r = 1 

a = 63° 26' and e = 116° 34'. 

In (fig. 144.) a = 1, m = }, 

24* 



282 



DIFFERENTIAL CALCULUS. [b. II. CH. XI. 



Multiple points. 



the origin as an union of three cusps, and the three values of 
t at this point are 

r = 60°, t = 180°, t = 300°. 

There are three other multiple points corresponding to 

9> — 0° or =360°, y = 120° or =480°, y = 240° or =600°, 

at each of which points we have 

r=l, a — 33° 41', and . == 146° 19'. 

In (fig. 145.) a = 1, m = ^, 

the origin is a cusp, and the tangent at this point is perpen- 
dicular to the axis. 

There are two other multiple points corresponding to 
<r= 90°, or =450°, y = 630° or =990°, 

at each of which points we have 

r — 1, s — 71° 34', and * = 108° 26'. 

In (fig. 146.) a = 1, M=z f, 

the origin is an union of five cusps, and the five values of r at 
this point are 

T = 18°, t — 90°, r = 162°, t — 234°, r — 306°. 

There are ten other multiple points ; five of these points cor- 
respond to 

^ = 90° or =450°, <p = 306° or =666°, 9 = 528° or =882° 
if =z 18° or = 738°, 9 = 234° or = 954° ; 

for each of these points 

r = 1.5, « = 46° 6', and * = 133° 54', 



^^^HM 



§ 174.] SINGULAR POINTS. 283 

Multiple points, 
the other five points correspond to 
^=198° or =558° cp — 4U° or =774°, ^ = 630° or =990°, 

cp = 126° or == 846°, <p =h 342° or = 1102°. 
for each of these points 

r = 0.5, e = 19° 6 ; and £ = 160° 54'. 

In (fig. 147.) a = l, m = i, 

the origin is a cusp, and the tangent at this point is perpen- 
dicular to the axis. 

There are four other multiple points corresponding to 

q> = 270° or = 630°, r == 9.045, \ — 86° 16' and b — 93° 44' 

<p=1170°or = 1530°, r = 0.955, *=58° 23' and ^ — 121° 37/ 

q>— 90°orz= 810°, r = 6.545, * = 81° 39' and a — 98° 21' 

q> — 990° or = 1710°, r = 0.345, s — 19° 58' and s — 160° 2' 

In (fig. 148.) a = 1, m = f , 

the origin is a union of three cusps, at which point the values 

of T are 

r = 90°, t — 210°, t — 330°. 

There are twelve other multiple points ; three of these points 
correspond to 

9 = 90° or =810°, q>=690° or == 1410°, <p = 210° or =1290° 

for each of these points 

r = 1.809, b — 78° 58' and • = 101° 2' ; 
three points correspond to 



284 DIFFERENTIAL CALCULUS. [b. II. CH. XL 

Multiple points. 

<p = 270° or = 630°, g> == 870° or = 1230°, 9 = 30° or =± 1470° 
for each of these points 

r = 1.309, e — 66° 27', and a = 113° 33'; 
three points correspond to 

<p =570° or = 930°, 9 =, 1170° or 1530°, v =330° or = 1770° 
for each of these points 

r = 0.691, i = 50° 27', and e = 129° 33' ; 
three points correspond to 

<p=: 390° or =1110°, 9 = 990° or = 1710°, <p = 510° or 1590° 

for each of these points 

r = 0.191, ■ = 28° 26', and £ = 151° 34'. 

When m is an even integer, or a fraction whose numerator is 
even, each cusp at the origin has another cusp opposite to it, 
which causes both of them to disappear ; and the origin, in- 
stead of being an union of cusps, is a multiple point where 
the curve has a contact or several contacts with itself. 

In (fig. 149.) a = 1, m = 2, 

the curve consists of two ovals, which have a contact at the 
origin, the value of T at this point is 

* = 135°. 

In (fig. 150.) a = 1, m = 4, 

the curve has two contacts with itself at the origin ; the two 
values of r at this point are 

t = 67° 30', r = 157° 30'. 



§ 174.] SINGULAR POINTS. 285 

Multiple points. 

In (fig. 151.) a = 1, m = §, 

the curve has a contact with itself at the origin, and consists 
of two distinct branches. The value of * at the origin is 

r = 45°. 

There are four other multiple points ; two of these points cor- 
respond to 

<P = 45° or == 765°, cp = 2:25° or = 585° 
for each of these points 

r = 1.5, « = 63° 57', and * = 111 3'; 
two points correspond to 

9 = 315° or = 1035°, 9" = 495° or = 855° 
for each of these points 

r = 0.5, . = 40° 54 , and ■ = 139° 6'. 

In (fig. 152.) a = 1 ■*=;•$» 

the curve has two contacts at the origin ; the two values of * 
at this point are 

t = 22° 30', r = 112° 30'. 

There are eight other multiple points ; four of these points 
correspond to 

<P = 22° 30' or = 382° 30', y = 292° 30 or = 652° 30^ 

9 = 562° 30 or — 922° 30', <f = 112° 30' or = 832° W 

for each of these points 

r = 1.5, b= 52° 25', and * = 127° 35', 

four points correspond to 



286 DIFFERENTIAL CALCULUS. [b. II. CH. XI. 

Multiple points. 

y =: 157°30' or = 517°30', <p=427°30' or z=787°30' 
9 — 697° 30' or = 1057° 30', 9 = 247° 30' or — 967° 30' 
for each of these points 

r = 0.5, a = 23° 25', and * = 156° 35'. 

In (fig. 153.) a = 1, m = f , 

the curve consists of two distinct branches, which are in con- 
tact at the origin ; the value of <p at this point is 

9 = 135°. 

There are eight other multiple points ; two of these points 
correspond to 

9 = 315° or = 1035°, 9 = 135° or = 1215° 

for each of these points 

r = 1.809, * = 82° 36', and « = 97° 24'; 
two points correspond to 

9 = 45° or = 405°, 9 == 945° or = 1305°, 
for each of these points 

r = 1.309, * = 73° 48', and e = 106° 12'; 
two points correspond to 

9 = 495° or =855°, 9 = 1395° or = 1755°, 
for each of these points 

r = 0.691, * = 61° 10', and « = 118° 50'; 
two points correspond to 

9 = 765° or = 1485°, 9 = 585° or = 1665°, 



i^^H^ 



§ 174] SINGULAR POINTS. 287 

Multiple points. 

for each of these points 

r = 0.191, ^ = 39° 5', and * = 140° 55'. 

When a is less than unity, the origin is a multiple point, and 
several branches of the curve stop at this point, if the negative 
values of the radius vector are neglected, while they continue 
through it if these values are retained. This example, there- 
fore, furnishes an analytic exception to the method of avoiding 
negative radii vectores given in B. I. § 45. In the following 
figures the dotted portions correspond to the negative radii 
vectores. 

In (fig. 154.) * — i, m= 1, 

the two values of r at the origin are 

T = 30°, r = 150°. 

In (fig. 155.) a = h m = 2, 

the two values of ? at the origin are 

T = 105°, t = 165°. 

Whether the dotted parts are included or not, the curve is 
continuous. 

In (fig. 156.) a = J, m = 3, 

the six values of r at the origin are 

t= 10°, t= 50°, r= 70°, 
t = 110°, ^130°, t = 170°. 

In (fig. 157.) a = %, m = 4, 

the four values of t at the origin are 



288 DIFFERENTIAL CALCULUS. [b. II. CH. XI. 

Multiple points. 

T = 52° 30', t = 82° 30', 
T = 142° 30', * = 172° 30'. 

In (fig. 158.) a = £, m — £, 

the two values of r at the origin are 

r == 60°, t = 120°. 
There is another multiple point corresponding to 

^ = 0° or == 360°, r = 0.5, t = 45° and = 135°, 
or 9=540°, r = — 0.5, r = 90°; 

the common tangent at this point is perpendicular to the axis. 

In (fig. 15&.) a = h m = f » 

the six values of t at the origin are 

r= 20°, t= 40°, r= 80°, 
r = 100°, t = 140°, r == 160°. 
There are three other multiple points, corresponding, respec- 
tively, to 
9 = o° or = 360°, <p = 12° or = 480°, <p = 240° or = 600° 

for each of which r = 0.5, 

or to 

9 = 180°, y = 420°, 9 = 660°, 

for each of which r = — 0.5 ; 

at each of these three points the values of s are 

E = 18° 26', b = 161° 34', and 8 = 90°. 

In (fig. 160.) a = J, m = £, 

the curve has a contact with itself at the origin, the tangent 



<§> 174] SINGULAR POINTS. 289 

Multiple points. 

at this point is perpendicular to the axis. There are three 
other multiple points ; one corresponds to 

<P = 90° = 450°, r = 0.5, ■ = 60°, and a = 120 ; 

the two others correspond to 

<P = 555° 48', c P — 1064° 12', r = 0.408, 

or to 

cp = 735° 48', tp = 883° 12', r = — 0.408, 

and at one of these two points, 

b = 129° 7', and e = 71° 8' ; 

at the other, 

£ = 50° 53', and e = 108° 52'. 

In (fig. 161.) a = t, m = f. 

The curve consists of two ovals and a continuous re-entering 
branch, which has a contact with itself and with each of the 
ovals at the origin ; the value of t at the origin is 

r ~ 45°. 

There are two other multiple points, corresponding to 

cp = 225° = 585°, y == 45° = 765° ; 
at each of these points 

r — 1, « — 60°, and e = 120°. 

In (fig. 162.) a = £, m = f. 

The curve has several contacts with itself at the origin ; the 
values of t at this point are 

r == 67° 30 r , t = 157° 30'. 
25 



290 DIFFERENTIAL CALCULUS. [b.II. CH. Xi. 

Multiple points. 

There are four other multiple points, corresponding to 
<P == 22° 30' or — 382° 30', 9 = 292° 30' or =652° 30' 
<9 = 562° 30' or =: 922° 30', y = 1 12° 30' or — 832° 30' 

at each of these points 

r = 1, « = 40° 54', and 8 = 139° 6', 

In (fig. 163.) a = i m = £. 

The two values of r at the origin are 

% = 60°, r=. 120°. 
There are four other multiple points ; one corresponds to 
g> = 0° or = 720°, r — 0.5, 
r — 63° 26' and t — 116° 34'; 
one point corresponds to 

cp — 180° or = 540°, r — 1.207, 
r — 81° 40' and r — 98° 20' ; 
one point corresponds to 

9 — 761° 4', r = 0.322, . = 127° 14', 
or to 

9 = 41° 4', r — — 0.322, . =5 66° 15' ; 
and one point corresponds to 

cp = 1218° 56', r = — 0.322, . = 113° 45', 
or to 

<p as 1398° 56', r = 0.322, * =• 52° 46'. 

In (fig. 164.) a = J, W'=?| 



<§> 174.] SINGULAR POINTS. 291 

Multiple points. 

The six values of * at the origin are 

r = 20°, r= 40°, r= 80°, 

r = 100°, t = 140°, * = 160°. 

There are fifteen other multiple points ; three correspond to 

<p = 0° or — 720°, q> = 240° or = 760°, <p = 480° or = 1200° 

r = J, ■ = 33° 41', and s = 146° 19' ; 

three correspond to 

¥ = 180° or=540°, ? = 660° or = 1020°, 9 = 60° or =1140, 

r = 1.207, ? = 66° 23' and a = 113° 37' ; 

three correspond to 

cp = 420° or = 880°, <p = 900° or = 1200°, <p = 300° or 1380° 

r = _ 0.207, = = 21° 31', and . = 15S° 29' ; 

three correspond to 

cp — 733° 41', cp == 1213° 41', 9 *= 253° 41', 

for each of which 

r dz 0.322, E = 156° 26', 

or to 

9 = 1273° 41', v == 313° 41', 9 = 793° 41'; 

for each of which 

r = — 0.322 « = 143 1'; 

three correspond to 

» = 406° 19' <p = 886° 19', g> =± 1366° 19'; 

for each of which 

r = — 0.322, » = 36° 59', 



292 DIFFERENTIAL CALCULUS. [b. II. CH. XI. 

Multiple points. 

or to 

<p =z 946° 19', <p = 1426° 19', <p = 826° 19; 
for each of which 

r = 0.322, e = 23° 34'. 

26. Construct the locus of the polar equation 

r — a<P -\- a—<P. 

Arts. This curve (fig. 165.) has an infinite number of mul- 
tiple points, corresponding to 

<p = ±: rc.l80°, 

in which n is any integer. 

175. When a curve is continuous at a point, but 
changes its direction so as to turn its curvature the op- 
posite way at this point, the point is called a point of 
contrary flexure, or a point of inflexion. 

Thus M (figs. 166 - 169) is such a point. 

176. Problem. To find the points of contrary flexure. 

Solution. It is evident, from the comparison of the two tan- 
gents M' T> (figs. 165-169.), and M" T" near M, 
that the value of the angle MTX or r is either a maximum 
or a minimum at the point M. 

The points of contrary flexure correspond, therefore, to 
the maxima and minima of the angle r. 

177. Corollary. When the equation of the curve is given in 
rectangular coordinates, we have by (549) 

tang, t — d e .y; 



§ 178.] SINGULAR POINTS. 293 

Multiple points. 

so that the maxima and minima of * correspond to those of 
d c . y, except at those points where r is a right angle. 

178. Corollary. It is evident, from (figs. 166-169), 
that the convexity of a curve is turned towards the axis 
of x. when the angle * (or its supplement, if the curve 
is below the axis) increases with the increase of x : 
otherwise the convexity is turned from the axis. 

179. Examples. 

1. Find the point of contrary flexure in the locus of exam- 
ple 1, § 164, and the tangent at this point. 

Solution. We have, in this case, 

d c .y = — [log. (x — a)]-a (x — a)-i 

dly = (z — of-? [log.(x — «)]-3 [log.(x-a) + 2]; 

so that the point of contrary flexure corresponds to the point 
M' (fig. 56.) for which 

log. (x — a) + 2 = 

x— a-\- 0.135. y — b — 0.5 
r — 61° 38'. 

2. Find the point of contrary flexure in the locus of example 
2, § 164, and the tangent at this point. 

Ans. It corresponds to 

x = a + 7.3S7, y ■= b + 3.694, r — 26° 33 . 

3. Find the point of contrary flexure in the locus of exam- 
ple 3, § 164, and the tangent at this point. 

25* 



294 DIFFERENTIAL CALCtfLUS. [b. II. CH« XI. 

Multiple points. 

Ans. It corresponds to 

x = a + 1, y — b + 1, r = 26° 33'. 

4. Find the point of contrary flexure in the locus of exam- 
ple 5, § 164, and the tangent at this point. 

Ans. It corresponds to 

x — a-\- 0.223, y = b — 0.335, r — 155° 57'. 

5. Find the point of contrary flexure in the locus of exam- 
ple 6, of § 164, and the tangent at this point. 

Ans. It corresponds to 

x = a + 0.340, y = b + 0.085, T = 135°. 

6. Find the points of contrary flexure in the locus of exam- 
ple 7, § 164. 

Ans. There are two which correspond to 

x = a + 0.683, y — 6 + 0.068, t=162°8' 

x = a + 0.073, y — b + 0.035, r— 31° 6'. 



<§> 180.] APPROXIMATION. 295 

Approximate value of an explicit function. 



CHAPTER XII. 

APPROXIMATION, 

Almost all theoretical results, when converted into 
numbers, are insusceptible of exact expression, and can 
only be obtained approximatively. Hence, in all its 
practical applications, ready and rapid means of ob- 
taining approximations are the only object of the exact 
science of mathematics ; and the great labor, which 
has been bestowed upon this subject, is the distinguish- 
ing characteristic of the modern science. 

ISO. Problem. To obtain by approximation the value 
of an explicit function. 

Solution. The only useful method of accomplishing this 
object is to arrange the function in a series of terms, which 
are susceptible of easy calculation and decrease as rapidly as 
possible. 

I. When the variable is very small, the function is, at once, 
arranged by means of AlacLaurin"s theorem (447) in a 
series of terms, which are multiplied by the successive powers 
of the variable, and are, therefore, usually decreasing. 

II. "When the values of the function and its differential 
coefficients are known for a given value of the variable ; the 
function can, for another value of the variable, which differs 
but little from the given one, be arranged, by means of Tay- 



296 DIFFERENTIAL CALCULUS. [b. II. CH. XII. 

Approximate value of an explicit function. 

lor's Theorem (445), according to the successive powers of the 
difference between the two values of the variable. 



III. Besides the formulas thus obtained, other formulas 
can often be found, by processes dependent upon the nature 
of the functions and the tact of the geometer ; and some for- 
mulas, often of great use, will be given in the Integral Calcu- 
lus. 

Scholium. Formulas (478, 484, 487, 492,493,500,501, 
504, 509, 513, 515), are examples of useful developments. 

181. Problem. To obtain, by approximation, the 
values of an implicit function, ivhen its value is known 
to differ but little from that of a given explicit func- 
tion. 

Solution. Let 

x =r the required implicit function 

t = the given explicit function 

x — t = e = the excess of x above t. 

Find the algebraic equation for determining e, and let it be 
reduced to the form 

e = Fx, 

where F x is a small function of x. which we may denote by 
a z, in which a is any small quantity, and z the function of x 
obtained by dividing e by a ; we have then, 

e = Fx = az (631) 

x=zt+e = t + Fx = t + az. (632) 



§ 180.] APPROXIMATION. 297 

Lagrange's theorem. 

Now we have by MacLaurin's Theorem for any function u of 
x if we develop it according to powers of a, and denote by w , 
d ca u , &c, the values of ud ca u, &c, when 

a = o 

a 2 a 3 

u = u + d c _ a u . a + df M u . — + d% u . jj^ + &c. 

Again, if we put (633) 

dL. u = u' } rf« u = u' , (634) 

we have by (566), 

d ct u = u' d ct x, . (635) 

d c . a u = u> d c . a x. (636) 

But the differentiation of (632), gives, by putting 

z' — d c . x z, (637) 

d c . a x — z + ad^z^z + az' d c _ a x, (638) 

whence 

d x- Z < 639 ) 

In the same way, the differential coefficient of (632), rela- 
tively to t, is 

«U * = 1 + a JJ z = 1 -f a z' d ct x, (640) 
whence 

d x - — L_ (641) 

" 1 — as" 

and, therefore, 

d c , a x— z d ct x (642) 

d ca u — u'z d ct x. (643) 



298 



DIFFERENTIAL CALCULUS. [b. II. CH. XII. 



Lagrange's theorem. 



The differential coefficient of this last equation, relatively to 
t, is 



or 



d\. a d c .t u = d ct (u' z d c , t x), 



d ca (u' d ct x) = d c A (u' z d cA x), 



(644) 



(645) 



in which any function whatever of x may be substituted for u' 
By substituting for u' in (645) the fnnction z n u' of z, we 
have 

d c . a (z n u' d cA x) = d c - t (%»+! u' d ct x). (646) 

Now the successive differential coefficients of (643), rela- 
tively to a, are by (646), 

dl a u == d ca (z u' d CJ x) = d cA (z 2 u' d ct x) (647) 

d* a u = d c , d c _ a (*» „/ d c ; t x) = d*, (»■ u> d CJ x) (648) 

and in general 

dl a u — d c . t d CM (zn-^u' d c _ t x) — d"- 1 (z n v! d c , t x) (649) 

Now if in (641 and 646) we take 
a = o 
we have 

d c a x =z 1 
d CM u = k' z 

t^'^-iK^) (650) 

whence by (631) 

a d c , a u = a u' z — u' JP. t (651 ) 



§ 180.] APPROXIMATION. 299 

Lagrange's theorem. 



«2 dl a u = d cA (a^ u' z*) = d cA [u' (F. ty] 
a" dl a u = d*- 1 (a* u' z n ) = d*-* [u' (F % t)*] (652) 
which, substituted in (633) give 

u = «„ + v! F.t + *•' [ "'j ( ^ )2 ] - + &e. . . . (653) 

which is called Lagrange's Theorem. 
Corollary* If 

u — x, (654) 

u' z=z 1, u Q = t 
and (650) becomes 

Corollary. If instead of (632), x had been the given func- 
tion of t -f- a % 

z =/.(* + a z) (656) 

we might have put 

x' — t + a z, (657) 

and u would have been a function of x', and that if such 

u = tp . x (658) 



we have 












and if we put 

< 


U— (p .f .x' 
= 4* 9 ,/- '. 








(659) 
(660) 


The formula (650) 


may be directly 


applied 


to this 


case. 





300 DIFFERENTIAL CALCULUS. [b. II. CH. XII. 

Lagrange's theorem. 

The theorem (65f£) under this form of application, has 
been often called Laplace's Theorem; but, regarding 
this change as obvious and insignificant, we do not 
hesitate to discard the latter name, and give the whole 
honor of the theorem to its true author Lagrange. 

Examples. 

1. Find the mth power of a root of the equation 

z = t + <*x? (661) 

in which « is a small quantity. 

Solution. In this case 

F.X = axP, F.t = a tP 



u =zt m , u' z= m t m ~ l 
dr- 1 [u' . (F, r) = d'j 1 (m a n r?+ m - *) 

=m«" {np-\-m — 1) (np-\-m— 2) (np+m — «+l) l n P+ m ~ n 

and, therefore, 

1 . z 

Corollary. When (662) 

m = 1 

(659) becomes (663) 

1 . Z 1 . Z . o 



§ 180.] APPROXIMATION. 301 

Lagrange's theorem. 

2. Find the value of x from the equation 

x __ f _|_ a e mx 

in which « is small and c is the Neperian base. 

Ans. (664) 

x — t -f ae mt + m a 2 e 2m '-j-—-9ffl'« 3 e zmt + &c. 

3. Find the value of e nx , from the preceding example. 

Ans. (665) 

e nr — e"' + « » e («+"> ' + £ " 2 n (2 m + w) c ( 2 ™+») f + &c. 



THE END. 



26 



ERRATA. 

Page 10, line 7, for polymonial read polynomial 1. 10, for b read d. — 1. 22, for 19 read 

20. 
Page 13, line 16 and 1. 19, for 2-4 read 25. 
Page 19, line 18, for A E read A B. 
Page 21, line 6, for quadrilateral read quadrilaterals. 
Page 22, line 7, for x' read z. 
Page 26, line 11, for sin. (<p + .#*) . sin. j2' read sin. (fp + •£') : sin. A'. — ]. 14, for 

c. J3C read c: £C. 
Page 27, line 8, for CB' + CD' read B' D' =— CB' + C'i>'. 
Page 30, liue — 4, for oftico axes read of the tico axes. 
Page 34, line — 4, for those read that. 
Page 39, line 7, for P' read B'. 
Page 42, line 21, for in plane read in a plane. 
Page 43, line 16, fur R' ' read L' '. 

Page 44, line 2, for^o?'«t Z) read pomi B. — 1. 4, for C read Z. — I. 7, for o?* read and. 
Page 45, line 13, for (x' — i') 2 -^ (#' — J/')' 2 read (z'— z)2 -J- (?/*— ?/)2. 
Page 46, line 7, for £'£ read ££. 
Page 47, lines 1 and 3, for AB read BB'. — h 14, for 82, read_84. — 1. 18, for the angles 

read the cosines of the angles. — 1. — 2, for angle read angles. 
Page 48, line 3, for 83 and 84 read 85 and 86. — 1. 6, for rcctanglar read rectangular. 

L — 5, for 84 read 86. 
Page 50, line 2, for 83 read 85. 

Page 64, line — 2, for — — read 

A c 

Pasre 65, line —4, fox AF — AC read AF — FC. 

Page 67, line 6, for CE = CE' = AF — AF' =A read AE = AE =s AF= AF' =. c. 
c A 

1. 8, for — - read 1. 13, for A2 — c2 read c2 — A®. 

'A c ' 

Page 69, line 8, for for ellipse read for the ellipse. 1. 16, for 3 read «• 

Paga 75, line 9, for P read R. 

Page 77, J. 10, for axes read axis. 

Page 78, line — 3 — 2 — 1, for 3 — a = ^ tv, (1 = ^ tv -\- a, COS. 3 = — sin. a 

read a — 3 = £ tt, (J =' a — ^ -r, cos. /J = sin. a. 
Page 79, line 1, for — read -f. 1. 19, for coordidatcs read coordinates. 
Page 80, line — 6,-5,-4, and — 2, for fi read /. • for v read u j for 7. read r. 
Page 96, line 1, for by read be. 

Page 107, line — 1, the first member of the equation should be doubled. 
Page 109, line — 7, dele may be. 1. — 6, dele would. 
Page 113, line 10, for E read — E. 
Page 1J4, line — 4, for S read X 

Page 117, line — 8, l'or 225 read 257 ; for example 2 read example 3. 
Page 118, line 2, for art. 23 read (23). 

Page 1]9, line 12, for lower read upper. 1. 15, for X = ± read % — -f- ^/ 
Page 120, line 7, for x read z. 1. 9, for y read y. ' 

2 2 

Page 124, line — 2 and page 125, 1. 4, for tang. I read — tan°- I 
Page 125, line 10, for CM' read CM. I. 11, for M and JIF read C and C. 
rage 126, line 12, dele — . 
Page 127, line — 4, for C read c. 

P Z™A ] A£rio S ia CwSSi £JY L 7 6 ' fm J M J™ C'Mandthe line drawn through 
m^Sre^^m^ lmC draWn thr ° USh °» L ~ 5 ' fw *' M ^i ?°r 

Page 130, line 1, for £ read F. 1. 10, for z , read to. 

Page 134, line 12, for zj. read z 2 . 

Page 136, line 3, for S read X 

Page 137, line - 2, for [(S . m >) 2 _ 2 ] read - • for S' read & 1. _ 1, for %' 4- y< 
read z'2 _|_ y 2. 2 \ J 2 



304 



p read + p. 1.— 4, for COS. 2 read 



1, for 771 



Page 138, line 11, for M read m. 1. —7, for 

cos. 2 for (Sm 1 sin. 2 « 2 )2 rea d (# mj sin.^aj) 2 ; i. 

read M. 
Page 144, line — 7, for y z read x z. 
Page 147, line 14, for eight read six. 
Page 149, line 3, for A read A% ; 1. 10, for t/te azis read £/ie axis of z. 

Page 153, line — 7, for x' read X ± . 

Page 155, line 6, for c read b. 

Page 157, line 8, for 366, read 365. 1. 9, for 373 read 374. 

Page 158, line — 5, for parameter read semi parameter. 

Page 165, line 7, for one, which contains only read a polynomial, which contains only positive, 

Page 171, line 9, for (p • n read y. n . 

Page 173, line 5, for 12 read 25. 

Page 180, line 1, for variable read functions . 

Page 181, line 4, dele second. 

Page 189, line 6, for vanished read vanished with the variable ; I. — 1, for ■#• read ^n. 

Page 191, line 3, for d2.f.f- read ^2 /f . 

Page 193, line — 4, for — read -4-. 

Page 194, line 1, for 483 read 473. 



x 7 



P^ a 04, li ne-6, f o r + - 28 . 4 . 5 . 6 . 7 "*- HJ.UM 

read 5 ; 1. — 2, for 81 read 82. 
Page 205, line 6, 8, and 12, for ^ 7V read £ n. 
Page 206, line 1, for + read — j 1. 2, for — read +. 
Page 209, line 4, for I read f . 
Page 214, line 1, for xO read x Q , 
Page 217, line 8, for 41 read 42. 
Page 218, line 1, for B read A. 
Page 219, line 1, for maximum read minimum. 
Page 220, line — 4 for A read M . 

Page 221, line 3, for MNMi read J^MM 1 . 1. — 6, for A read Z . 
Page 222, line — 4, dele t/te o?-cZ£t- of. 
Page 225, line — 9, for lo read J 2 . 
Page 225, line 12, for 548 read 550. 
Page 228, line — 1, for C read c. 

Page 229, line — 10, for 2 cos. £ cp read 2 sin. 2 £ 9 ; 1. — 7, for 39 read 37. 
Page 230, line 12. for cos. read sin. 
Page 233, line 5, for R read r. 
Page 237, line 5, for or roadawd. 
Page 238, line 3, for 128 read 123. 
Page 239, line — 4, for x y read x z. 
Page 241, line 5, for sin. 1 read sin. 2 . 

Page 242, line 6, for sin read cos. ; 1. — 1 for 2 (2 7T read 2 7T (2. 
Page 243, line 5, for curvature read curve. 
Page 245, line 2, for S read s |. 

Page 246, line 4, for 3 read |. 

Page 249, line 8, for drawn read o/a straight line drawn. 

Page 254, line 9, for tangents read tangent ; 1. 10, for # read y. 

Page 256, line 12, for reprr.nts read represents. 

Page 260, line 7, for junction read function ; 1. 6, for zero read M7iit7/, 

Page 266, line — 3, for increasing read decreasing. 

Page 266, line 6, for M' lead J)f j 1. 14, for P read P". 

Page 269, line — 1, for p' 2 T 2 r ead T 2 P 2 > 

Page 271, line 14, and page 274, 1. —9, for second read first. 
Page 288, dele line 9. 



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